Check whether the 5.0 N force is enough to put the puck into motion.

Question

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Answer 1

Question

Chapter 3, Problem 61P

(a)

To determine

Check whether the 5.0 N force is enough to put the puck into motion.

(a)

Expert Solution

Answer to Problem 61P

5.0 N force is not enough to put the puck into motion.

Explanation of Solution

Puck will start to move if the applied force is greater than the static frictional force. Write the equation to find the static frictional force.

$fst=μsmg$

Here, $fst$ is the static frictional force, $μs$ is the coefficient of static friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.35$ for $μs$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fst$ .

$fst=(0.35)(2.0 kg)(9.8 m/s2)=6.86 N$

From the calculation, it is evident that the applied force 5.0 N is lesser than that of the static frictional force. So, the puck cannot move.

Thus, 5.0 N force is not enough to put the puck into motion.

(b)

To determine

Check whether the sudden increase in applied force to 7.5 N force while pushing can put the puck into motion or not.

(b)

Expert Solution

Answer to Problem 61P

The 7.5 N force can put the puck into motion.

Explanation of Solution

Puck will start to move if the force is greater than the kinetic frictional force. Write the equation to find the kinetic frictional force.

$fk=μkmg$ (I)

Here, $fk$ is the kinetic frictional force, $μk$ is the coefficient of kinetic friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(9.8 m/s2)=4.9 N$

From the calculation, it is evident that the applied force 7.5 N is greater than that of the kinetic frictional force. So, the puck will move.

Thus, the 7.5 N force can put the puck into motion.

(c)

To determine

The acceleration of puck.

(c)

Expert Solution

Answer to Problem 61P

Acceleration will be $0.55 m/s2$ .

Explanation of Solution

Write the relation between applied force, friction force while puck is moving and the acceleration.

$F−fk=ma$

Here, $F$ is the applied force and $a$ is the acceleration.

Conclusion:

Substitute $6.0 N$ for $F$ , $4.9 N$ for $f$ and $2.0 kg$ for $m$ in the above equation to find $a$ .

$6.0 N−4.9 N=(2.0 kg)a1.1 N=(2.0 kg)aa=1.1 N2.0 kg=0.55 m/s2$

Thus, the acceleration will be $0.55 m/s2$ .

(d)

To determine

The acceleration of puck in moon and state the reason for difference, if any.

(d)

Expert Solution

Answer to Problem 61P

Acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Explanation of Solution

Write the equation to find the kinetic frictional force in moon.

$fk=μkmgmoon$

Here, $gmoon$ is the gravitational acceleration in moon.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $1.6m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(1.6 m/s2)=0.8 N$

Substitute $6.0 N$ for $F$ , $0.8 N$ for $fk$ and $2.0 kg$ for $m$ in equation (I) to find $a$ .

$6.0 N−0.8 N=(2.0 kg)a5.2 N=(2.0 kg)aa=5.2 N2.0 kg=2.6 m/s2$

The acceleration of puck with same applied force in moon is much higher than that on Earth. The reason is that the acceleration due to gravity on moon is only 1/6^th of Earth. It results in less frictional force on moon. As a result, acceleration will be more.

Thus, the acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

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Answer 2

Question

Chapter 3, Problem 61P

(a)

To determine

Check whether the 5.0 N force is enough to put the puck into motion.

(a)

Expert Solution

Answer to Problem 61P

5.0 N force is not enough to put the puck into motion.

Explanation of Solution

Puck will start to move if the applied force is greater than the static frictional force. Write the equation to find the static frictional force.

$fst=μsmg$

Here, $fst$ is the static frictional force, $μs$ is the coefficient of static friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.35$ for $μs$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fst$ .

$fst=(0.35)(2.0 kg)(9.8 m/s2)=6.86 N$

From the calculation, it is evident that the applied force 5.0 N is lesser than that of the static frictional force. So, the puck cannot move.

Thus, 5.0 N force is not enough to put the puck into motion.

(b)

To determine

Check whether the sudden increase in applied force to 7.5 N force while pushing can put the puck into motion or not.

(b)

Expert Solution

Answer to Problem 61P

The 7.5 N force can put the puck into motion.

Explanation of Solution

Puck will start to move if the force is greater than the kinetic frictional force. Write the equation to find the kinetic frictional force.

$fk=μkmg$ (I)

Here, $fk$ is the kinetic frictional force, $μk$ is the coefficient of kinetic friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(9.8 m/s2)=4.9 N$

From the calculation, it is evident that the applied force 7.5 N is greater than that of the kinetic frictional force. So, the puck will move.

Thus, the 7.5 N force can put the puck into motion.

(c)

To determine

The acceleration of puck.

(c)

Expert Solution

Answer to Problem 61P

Acceleration will be $0.55 m/s2$ .

Explanation of Solution

Write the relation between applied force, friction force while puck is moving and the acceleration.

$F−fk=ma$

Here, $F$ is the applied force and $a$ is the acceleration.

Conclusion:

Substitute $6.0 N$ for $F$ , $4.9 N$ for $f$ and $2.0 kg$ for $m$ in the above equation to find $a$ .

$6.0 N−4.9 N=(2.0 kg)a1.1 N=(2.0 kg)aa=1.1 N2.0 kg=0.55 m/s2$

Thus, the acceleration will be $0.55 m/s2$ .

(d)

To determine

The acceleration of puck in moon and state the reason for difference, if any.

(d)

Expert Solution

Answer to Problem 61P

Acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Explanation of Solution

Write the equation to find the kinetic frictional force in moon.

$fk=μkmgmoon$

Here, $gmoon$ is the gravitational acceleration in moon.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $1.6m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(1.6 m/s2)=0.8 N$

Substitute $6.0 N$ for $F$ , $0.8 N$ for $fk$ and $2.0 kg$ for $m$ in equation (I) to find $a$ .

$6.0 N−0.8 N=(2.0 kg)a5.2 N=(2.0 kg)aa=5.2 N2.0 kg=2.6 m/s2$

The acceleration of puck with same applied force in moon is much higher than that on Earth. The reason is that the acceleration due to gravity on moon is only 1/6^th of Earth. It results in less frictional force on moon. As a result, acceleration will be more.

Thus, the acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

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Answer 3

Question

Chapter 3, Problem 61P

(a)

To determine

Check whether the 5.0 N force is enough to put the puck into motion.

(a)

Expert Solution

Answer to Problem 61P

5.0 N force is not enough to put the puck into motion.

Explanation of Solution

Puck will start to move if the applied force is greater than the static frictional force. Write the equation to find the static frictional force.

$fst=μsmg$

Here, $fst$ is the static frictional force, $μs$ is the coefficient of static friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.35$ for $μs$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fst$ .

$fst=(0.35)(2.0 kg)(9.8 m/s2)=6.86 N$

From the calculation, it is evident that the applied force 5.0 N is lesser than that of the static frictional force. So, the puck cannot move.

Thus, 5.0 N force is not enough to put the puck into motion.

(b)

To determine

Check whether the sudden increase in applied force to 7.5 N force while pushing can put the puck into motion or not.

(b)

Expert Solution

Answer to Problem 61P

The 7.5 N force can put the puck into motion.

Explanation of Solution

Puck will start to move if the force is greater than the kinetic frictional force. Write the equation to find the kinetic frictional force.

$fk=μkmg$ (I)

Here, $fk$ is the kinetic frictional force, $μk$ is the coefficient of kinetic friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(9.8 m/s2)=4.9 N$

From the calculation, it is evident that the applied force 7.5 N is greater than that of the kinetic frictional force. So, the puck will move.

Thus, the 7.5 N force can put the puck into motion.

(c)

To determine

The acceleration of puck.

(c)

Expert Solution

Answer to Problem 61P

Acceleration will be $0.55 m/s2$ .

Explanation of Solution

Write the relation between applied force, friction force while puck is moving and the acceleration.

$F−fk=ma$

Here, $F$ is the applied force and $a$ is the acceleration.

Conclusion:

Substitute $6.0 N$ for $F$ , $4.9 N$ for $f$ and $2.0 kg$ for $m$ in the above equation to find $a$ .

$6.0 N−4.9 N=(2.0 kg)a1.1 N=(2.0 kg)aa=1.1 N2.0 kg=0.55 m/s2$

Thus, the acceleration will be $0.55 m/s2$ .

(d)

To determine

The acceleration of puck in moon and state the reason for difference, if any.

(d)

Expert Solution

Answer to Problem 61P

Acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Explanation of Solution

Write the equation to find the kinetic frictional force in moon.

$fk=μkmgmoon$

Here, $gmoon$ is the gravitational acceleration in moon.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $1.6m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(1.6 m/s2)=0.8 N$

Substitute $6.0 N$ for $F$ , $0.8 N$ for $fk$ and $2.0 kg$ for $m$ in equation (I) to find $a$ .

$6.0 N−0.8 N=(2.0 kg)a5.2 N=(2.0 kg)aa=5.2 N2.0 kg=2.6 m/s2$

The acceleration of puck with same applied force in moon is much higher than that on Earth. The reason is that the acceleration due to gravity on moon is only 1/6^th of Earth. It results in less frictional force on moon. As a result, acceleration will be more.

Thus, the acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

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Answer 4

Question

Chapter 3, Problem 61P

(a)

To determine

Check whether the 5.0 N force is enough to put the puck into motion.

(a)

Expert Solution

Answer to Problem 61P

5.0 N force is not enough to put the puck into motion.

Explanation of Solution

Puck will start to move if the applied force is greater than the static frictional force. Write the equation to find the static frictional force.

$fst=μsmg$

Here, $fst$ is the static frictional force, $μs$ is the coefficient of static friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.35$ for $μs$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fst$ .

$fst=(0.35)(2.0 kg)(9.8 m/s2)=6.86 N$

From the calculation, it is evident that the applied force 5.0 N is lesser than that of the static frictional force. So, the puck cannot move.

Thus, 5.0 N force is not enough to put the puck into motion.

(b)

To determine

Check whether the sudden increase in applied force to 7.5 N force while pushing can put the puck into motion or not.

(b)

Expert Solution

Answer to Problem 61P

The 7.5 N force can put the puck into motion.

Explanation of Solution

Puck will start to move if the force is greater than the kinetic frictional force. Write the equation to find the kinetic frictional force.

$fk=μkmg$ (I)

Here, $fk$ is the kinetic frictional force, $μk$ is the coefficient of kinetic friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(9.8 m/s2)=4.9 N$

From the calculation, it is evident that the applied force 7.5 N is greater than that of the kinetic frictional force. So, the puck will move.

Thus, the 7.5 N force can put the puck into motion.

(c)

To determine

The acceleration of puck.

(c)

Expert Solution

Answer to Problem 61P

Acceleration will be $0.55 m/s2$ .

Explanation of Solution

Write the relation between applied force, friction force while puck is moving and the acceleration.

$F−fk=ma$

Here, $F$ is the applied force and $a$ is the acceleration.

Conclusion:

Substitute $6.0 N$ for $F$ , $4.9 N$ for $f$ and $2.0 kg$ for $m$ in the above equation to find $a$ .

$6.0 N−4.9 N=(2.0 kg)a1.1 N=(2.0 kg)aa=1.1 N2.0 kg=0.55 m/s2$

Thus, the acceleration will be $0.55 m/s2$ .

(d)

To determine

The acceleration of puck in moon and state the reason for difference, if any.

(d)

Expert Solution

Answer to Problem 61P

Acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Explanation of Solution

Write the equation to find the kinetic frictional force in moon.

$fk=μkmgmoon$

Here, $gmoon$ is the gravitational acceleration in moon.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $1.6m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(1.6 m/s2)=0.8 N$

Substitute $6.0 N$ for $F$ , $0.8 N$ for $fk$ and $2.0 kg$ for $m$ in equation (I) to find $a$ .

$6.0 N−0.8 N=(2.0 kg)a5.2 N=(2.0 kg)aa=5.2 N2.0 kg=2.6 m/s2$

The acceleration of puck with same applied force in moon is much higher than that on Earth. The reason is that the acceleration due to gravity on moon is only 1/6^th of Earth. It results in less frictional force on moon. As a result, acceleration will be more.

Thus, the acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 3, Problem 61P

(a)

To determine

Check whether the 5.0 N force is enough to put the puck into motion.

(a)

Expert Solution

Answer to Problem 61P

5.0 N force is not enough to put the puck into motion.

Explanation of Solution

Puck will start to move if the applied force is greater than the static frictional force. Write the equation to find the static frictional force.

$fst=μsmg$

Here, $fst$ is the static frictional force, $μs$ is the coefficient of static friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.35$ for $μs$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fst$ .

$fst=(0.35)(2.0 kg)(9.8 m/s2)=6.86 N$

From the calculation, it is evident that the applied force 5.0 N is lesser than that of the static frictional force. So, the puck cannot move.

Thus, 5.0 N force is not enough to put the puck into motion.

(b)

To determine

Check whether the sudden increase in applied force to 7.5 N force while pushing can put the puck into motion or not.

(b)

Expert Solution

Answer to Problem 61P

The 7.5 N force can put the puck into motion.

Explanation of Solution

Puck will start to move if the force is greater than the kinetic frictional force. Write the equation to find the kinetic frictional force.

$fk=μkmg$ (I)

Here, $fk$ is the kinetic frictional force, $μk$ is the coefficient of kinetic friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(9.8 m/s2)=4.9 N$

From the calculation, it is evident that the applied force 7.5 N is greater than that of the kinetic frictional force. So, the puck will move.

Thus, the 7.5 N force can put the puck into motion.

(c)

To determine

The acceleration of puck.

(c)

Expert Solution

Answer to Problem 61P

Acceleration will be $0.55 m/s2$ .

Explanation of Solution

Write the relation between applied force, friction force while puck is moving and the acceleration.

$F−fk=ma$

Here, $F$ is the applied force and $a$ is the acceleration.

Conclusion:

Substitute $6.0 N$ for $F$ , $4.9 N$ for $f$ and $2.0 kg$ for $m$ in the above equation to find $a$ .

$6.0 N−4.9 N=(2.0 kg)a1.1 N=(2.0 kg)aa=1.1 N2.0 kg=0.55 m/s2$

Thus, the acceleration will be $0.55 m/s2$ .

(d)

To determine

The acceleration of puck in moon and state the reason for difference, if any.

(d)

Expert Solution

Answer to Problem 61P

Acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Explanation of Solution

Write the equation to find the kinetic frictional force in moon.

$fk=μkmgmoon$

Here, $gmoon$ is the gravitational acceleration in moon.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $1.6m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(1.6 m/s2)=0.8 N$

Substitute $6.0 N$ for $F$ , $0.8 N$ for $fk$ and $2.0 kg$ for $m$ in equation (I) to find $a$ .

$6.0 N−0.8 N=(2.0 kg)a5.2 N=(2.0 kg)aa=5.2 N2.0 kg=2.6 m/s2$

The acceleration of puck with same applied force in moon is much higher than that on Earth. The reason is that the acceleration due to gravity on moon is only 1/6^th of Earth. It results in less frictional force on moon. As a result, acceleration will be more.

Thus, the acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Answer 6

Chapter 3, Problem 61P

(a)

To determine

Check whether the 5.0 N force is enough to put the puck into motion.

(a)

Expert Solution

Answer to Problem 61P

5.0 N force is not enough to put the puck into motion.

Explanation of Solution

Puck will start to move if the applied force is greater than the static frictional force. Write the equation to find the static frictional force.

$fst=μsmg$

Here, $fst$ is the static frictional force, $μs$ is the coefficient of static friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.35$ for $μs$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fst$ .

$fst=(0.35)(2.0 kg)(9.8 m/s2)=6.86 N$

From the calculation, it is evident that the applied force 5.0 N is lesser than that of the static frictional force. So, the puck cannot move.

Thus, 5.0 N force is not enough to put the puck into motion.

Answer 7

Chapter 3, Problem 61P

(a)

To determine

Check whether the 5.0 N force is enough to put the puck into motion.

Answer 8

(a)

Expert Solution

Answer to Problem 61P

5.0 N force is not enough to put the puck into motion.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Puck will start to move if the applied force is greater than the static frictional force. Write the equation to find the static frictional force.

$fst=μsmg$

Here, $fst$ is the static frictional force, $μs$ is the coefficient of static friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.35$ for $μs$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fst$ .

$fst=(0.35)(2.0 kg)(9.8 m/s2)=6.86 N$

From the calculation, it is evident that the applied force 5.0 N is lesser than that of the static frictional force. So, the puck cannot move.

Thus, 5.0 N force is not enough to put the puck into motion.

Answer 13

(b)

To determine

Check whether the sudden increase in applied force to 7.5 N force while pushing can put the puck into motion or not.

(b)

Expert Solution

Answer to Problem 61P

The 7.5 N force can put the puck into motion.

Explanation of Solution

Puck will start to move if the force is greater than the kinetic frictional force. Write the equation to find the kinetic frictional force.

$fk=μkmg$ (I)

Here, $fk$ is the kinetic frictional force, $μk$ is the coefficient of kinetic friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(9.8 m/s2)=4.9 N$

From the calculation, it is evident that the applied force 7.5 N is greater than that of the kinetic frictional force. So, the puck will move.

Thus, the 7.5 N force can put the puck into motion.

Answer 14

(b)

To determine

Check whether the sudden increase in applied force to 7.5 N force while pushing can put the puck into motion or not.

Answer 15

(b)

Expert Solution

Answer to Problem 61P

The 7.5 N force can put the puck into motion.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Puck will start to move if the force is greater than the kinetic frictional force. Write the equation to find the kinetic frictional force.

$fk=μkmg$ (I)

Here, $fk$ is the kinetic frictional force, $μk$ is the coefficient of kinetic friction, $m$ is the mass of puck and $g$ is the gravitational acceleration.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $9.8 m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(9.8 m/s2)=4.9 N$

From the calculation, it is evident that the applied force 7.5 N is greater than that of the kinetic frictional force. So, the puck will move.

Thus, the 7.5 N force can put the puck into motion.

Answer 20

(c)

To determine

The acceleration of puck.

(c)

Expert Solution

Answer to Problem 61P

Acceleration will be $0.55 m/s2$ .

Explanation of Solution

Write the relation between applied force, friction force while puck is moving and the acceleration.

$F−fk=ma$

Here, $F$ is the applied force and $a$ is the acceleration.

Conclusion:

Substitute $6.0 N$ for $F$ , $4.9 N$ for $f$ and $2.0 kg$ for $m$ in the above equation to find $a$ .

$6.0 N−4.9 N=(2.0 kg)a1.1 N=(2.0 kg)aa=1.1 N2.0 kg=0.55 m/s2$

Thus, the acceleration will be $0.55 m/s2$ .

Answer 21

(c)

To determine

The acceleration of puck.

Answer 22

(c)

Expert Solution

Answer to Problem 61P

Acceleration will be $0.55 m/s2$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the relation between applied force, friction force while puck is moving and the acceleration.

$F−fk=ma$

Here, $F$ is the applied force and $a$ is the acceleration.

Conclusion:

Substitute $6.0 N$ for $F$ , $4.9 N$ for $f$ and $2.0 kg$ for $m$ in the above equation to find $a$ .

$6.0 N−4.9 N=(2.0 kg)a1.1 N=(2.0 kg)aa=1.1 N2.0 kg=0.55 m/s2$

Thus, the acceleration will be $0.55 m/s2$ .

Answer 27

(d)

To determine

The acceleration of puck in moon and state the reason for difference, if any.

(d)

Expert Solution

Answer to Problem 61P

Acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Explanation of Solution

Write the equation to find the kinetic frictional force in moon.

$fk=μkmgmoon$

Here, $gmoon$ is the gravitational acceleration in moon.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $1.6m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(1.6 m/s2)=0.8 N$

Substitute $6.0 N$ for $F$ , $0.8 N$ for $fk$ and $2.0 kg$ for $m$ in equation (I) to find $a$ .

$6.0 N−0.8 N=(2.0 kg)a5.2 N=(2.0 kg)aa=5.2 N2.0 kg=2.6 m/s2$

The acceleration of puck with same applied force in moon is much higher than that on Earth. The reason is that the acceleration due to gravity on moon is only 1/6^th of Earth. It results in less frictional force on moon. As a result, acceleration will be more.

Thus, the acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Answer 28

(d)

To determine

The acceleration of puck in moon and state the reason for difference, if any.

Answer 29

(d)

Expert Solution

Answer to Problem 61P

Acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the equation to find the kinetic frictional force in moon.

$fk=μkmgmoon$

Here, $gmoon$ is the gravitational acceleration in moon.

Conclusion:

Substitute $0.25$ for $μk$ , $2.0 kg$ for $m$ and $1.6m/s2$ for $g$ in the above equation to find $fk$ .

$fk=(0.25)(2.0 kg)(1.6 m/s2)=0.8 N$

Substitute $6.0 N$ for $F$ , $0.8 N$ for $fk$ and $2.0 kg$ for $m$ in equation (I) to find $a$ .

$6.0 N−0.8 N=(2.0 kg)a5.2 N=(2.0 kg)aa=5.2 N2.0 kg=2.6 m/s2$

The acceleration of puck with same applied force in moon is much higher than that on Earth. The reason is that the acceleration due to gravity on moon is only 1/6^th of Earth. It results in less frictional force on moon. As a result, acceleration will be more.

Thus, the acceleration will be $2.6 m/s2$ which is higher than that in earth and the reason is that the lesser value of gravitational acceleration on moon creates only less frictional force.