The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 3, Problem 3.1P

To determine

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Expert Solution & Answer

Answer to Problem 3.1P

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Explanation of Solution

Given:

$m=2$ , $a=g$ , $s=h$ and $u=0$

The height of the platform from the ground is 10 feet.

The height of the point from where the mass falls is 30 feet.

Concept Used:

Following kinematics equations are employed for evaluating the speed and height as the functions of time:

$v=u+ats=ut+12at2$

The latter equation is used for calculating the time taken by the mass to reach the platform and ground.

Calculation:

Since,

$v=u+at$

Therefore, on keeping the values for $u$ and $a$ .

$v(t)=gt2$

Similarly, for the height, that is

Since,

$s=ut+12at2$

On substituting the values

$h(t)=12gt2$

The time taken to reach the platform:

$h(t)=ut+12gt2⇒6.097=0⋅t+12⋅9.8⋅t2⇒t2=12.1959.81=1⇒t=1sec$

Similarly, The time taken to reach the ground:

$h(t)=ut+12gt2⇒9.146=0⋅t+12⋅9.8⋅t2⇒t2=18.299.81=1.86⇒t=1.36sec$ .

Conclusion:

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Want to see more full solutions like this?

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

Students have asked these similar questions

Show work on how to obtain P2 and T2. If using any table, please refer to it. If applying interpolation method, please show the work.

cast-iron roller FIGURE P11-3 Shaft Design for Problems 11-17 Chapter 11 BEARINGS AND LUBRICATION 677 gear key P assume bearings act as simple supports 11-18 Problem 7-18 determined the half-width of the contact patch for a 1.575-in-dia steel cylinder, 9.843 in long, rolled against a flat aluminum plate with 900 lb of force to be 0.0064 in. If the cylinder rolls at 800 rpm, determine its lubrication condition with ISO VG 1000 oil at 200°F. R₁ = 64 μin (cylinder); R₁ = 32 μin (plate). 11-19 The shaft shown in Figure P11-4 was designed in Problem 10-19. For the data in the row(s) assigned from Table P11-1, and the corresponding diameter of shaft found in Problem 10-19, design suitable bearings to support the load for at least 5E8 cycles at 1200 rpm. State all assumptions. (a) (b) Using hydrodynamically lubricated bronze sleeve bearings with ON = 40, 1/ d=0.80, and a clearance ratio of 0.002 5. Using deep-groove ball bearings for a 10% failure rate. *11-20 Problem 7-20 determined the…

Calculate the shear force at the point D on the beam below. Take F=19 and remember that this quantity is to be used to calculate both forces and lengths. 15F A с

Answer 2

Question

Chapter 3, Problem 3.1P

To determine

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Expert Solution & Answer

Answer to Problem 3.1P

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Explanation of Solution

Given:

$m=2$ , $a=g$ , $s=h$ and $u=0$

The height of the platform from the ground is 10 feet.

The height of the point from where the mass falls is 30 feet.

Concept Used:

Following kinematics equations are employed for evaluating the speed and height as the functions of time:

$v=u+ats=ut+12at2$

The latter equation is used for calculating the time taken by the mass to reach the platform and ground.

Calculation:

Since,

$v=u+at$

Therefore, on keeping the values for $u$ and $a$ .

$v(t)=gt2$

Similarly, for the height, that is

Since,

$s=ut+12at2$

On substituting the values

$h(t)=12gt2$

The time taken to reach the platform:

$h(t)=ut+12gt2⇒6.097=0⋅t+12⋅9.8⋅t2⇒t2=12.1959.81=1⇒t=1sec$

Similarly, The time taken to reach the ground:

$h(t)=ut+12gt2⇒9.146=0⋅t+12⋅9.8⋅t2⇒t2=18.299.81=1.86⇒t=1.36sec$ .

Conclusion:

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Want to see more full solutions like this?

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

Answer 3

Question

Chapter 3, Problem 3.1P

To determine

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Expert Solution & Answer

Answer to Problem 3.1P

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Explanation of Solution

Given:

$m=2$ , $a=g$ , $s=h$ and $u=0$

The height of the platform from the ground is 10 feet.

The height of the point from where the mass falls is 30 feet.

Concept Used:

Following kinematics equations are employed for evaluating the speed and height as the functions of time:

$v=u+ats=ut+12at2$

The latter equation is used for calculating the time taken by the mass to reach the platform and ground.

Calculation:

Since,

$v=u+at$

Therefore, on keeping the values for $u$ and $a$ .

$v(t)=gt2$

Similarly, for the height, that is

Since,

$s=ut+12at2$

On substituting the values

$h(t)=12gt2$

The time taken to reach the platform:

$h(t)=ut+12gt2⇒6.097=0⋅t+12⋅9.8⋅t2⇒t2=12.1959.81=1⇒t=1sec$

Similarly, The time taken to reach the ground:

$h(t)=ut+12gt2⇒9.146=0⋅t+12⋅9.8⋅t2⇒t2=18.299.81=1.86⇒t=1.36sec$ .

Conclusion:

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Want to see more full solutions like this?

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

Answer 4

Question

Chapter 3, Problem 3.1P

To determine

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Expert Solution & Answer

Answer to Problem 3.1P

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Explanation of Solution

Given:

$m=2$ , $a=g$ , $s=h$ and $u=0$

The height of the platform from the ground is 10 feet.

The height of the point from where the mass falls is 30 feet.

Concept Used:

Following kinematics equations are employed for evaluating the speed and height as the functions of time:

$v=u+ats=ut+12at2$

The latter equation is used for calculating the time taken by the mass to reach the platform and ground.

Calculation:

Since,

$v=u+at$

Therefore, on keeping the values for $u$ and $a$ .

$v(t)=gt2$

Similarly, for the height, that is

Since,

$s=ut+12at2$

On substituting the values

$h(t)=12gt2$

The time taken to reach the platform:

$h(t)=ut+12gt2⇒6.097=0⋅t+12⋅9.8⋅t2⇒t2=12.1959.81=1⇒t=1sec$

Similarly, The time taken to reach the ground:

$h(t)=ut+12gt2⇒9.146=0⋅t+12⋅9.8⋅t2⇒t2=18.299.81=1.86⇒t=1.36sec$ .

Conclusion:

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Want to see more full solutions like this?

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

Answer 5

Chapter 3, Problem 3.1P

To determine

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Expert Solution & Answer

Answer to Problem 3.1P

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Explanation of Solution

Given:

$m=2$ , $a=g$ , $s=h$ and $u=0$

The height of the platform from the ground is 10 feet.

The height of the point from where the mass falls is 30 feet.

Concept Used:

Following kinematics equations are employed for evaluating the speed and height as the functions of time:

$v=u+ats=ut+12at2$

The latter equation is used for calculating the time taken by the mass to reach the platform and ground.

Calculation:

Since,

$v=u+at$

Therefore, on keeping the values for $u$ and $a$ .

$v(t)=gt2$

Similarly, for the height, that is

Since,

$s=ut+12at2$

On substituting the values

$h(t)=12gt2$

The time taken to reach the platform:

$h(t)=ut+12gt2⇒6.097=0⋅t+12⋅9.8⋅t2⇒t2=12.1959.81=1⇒t=1sec$

Similarly, The time taken to reach the ground:

$h(t)=ut+12gt2⇒9.146=0⋅t+12⋅9.8⋅t2⇒t2=18.299.81=1.86⇒t=1.36sec$ .

Conclusion:

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Answer 6

Chapter 3, Problem 3.1P

To determine

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Expert Solution & Answer

Answer to Problem 3.1P

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Explanation of Solution

Given:

$m=2$ , $a=g$ , $s=h$ and $u=0$

The height of the platform from the ground is 10 feet.

The height of the point from where the mass falls is 30 feet.

Concept Used:

Following kinematics equations are employed for evaluating the speed and height as the functions of time:

$v=u+ats=ut+12at2$

The latter equation is used for calculating the time taken by the mass to reach the platform and ground.

Calculation:

Since,

$v=u+at$

Therefore, on keeping the values for $u$ and $a$ .

$v(t)=gt2$

Similarly, for the height, that is

Since,

$s=ut+12at2$

On substituting the values

$h(t)=12gt2$

The time taken to reach the platform:

$h(t)=ut+12gt2⇒6.097=0⋅t+12⋅9.8⋅t2⇒t2=12.1959.81=1⇒t=1sec$

Similarly, The time taken to reach the ground:

$h(t)=ut+12gt2⇒9.146=0⋅t+12⋅9.8⋅t2⇒t2=18.299.81=1.86⇒t=1.36sec$ .

Conclusion:

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Answer 7

Chapter 3, Problem 3.1P

To determine

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Answer 8

Expert Solution & Answer

Answer to Problem 3.1P

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

$m=2$ , $a=g$ , $s=h$ and $u=0$

The height of the platform from the ground is 10 feet.

The height of the point from where the mass falls is 30 feet.

Concept Used:

Following kinematics equations are employed for evaluating the speed and height as the functions of time:

$v=u+ats=ut+12at2$

The latter equation is used for calculating the time taken by the mass to reach the platform and ground.

Calculation:

Since,

$v=u+at$

Therefore, on keeping the values for $u$ and $a$ .

$v(t)=gt2$

Similarly, for the height, that is

Since,

$s=ut+12at2$

On substituting the values

$h(t)=12gt2$

The time taken to reach the platform:

$h(t)=ut+12gt2⇒6.097=0⋅t+12⋅9.8⋅t2⇒t2=12.1959.81=1⇒t=1sec$

Similarly, The time taken to reach the ground:

$h(t)=ut+12gt2⇒9.146=0⋅t+12⋅9.8⋅t2⇒t2=18.299.81=1.86⇒t=1.36sec$ .

Conclusion:

The speed and height of the mass are as:

$v(t)=gt2h(t)=12gt2$

The time taken by the mass to reach the platform and ground is 1 second and 1.36 second respectively.

Answer 12

Ch. 3 - Prob. 3.1P Ch. 3 - A baseball is thrown horizontally from the...Ch. 3 - For the mass shown in Figure 3.1.3b. m=10 kg, =25...Ch. 3 - A particle of mass m=19 kg slides down a...Ch. 3 - A particle of mass m slides down a frictionless...Ch. 3 - A radar tracks the flight of a projectile (see...Ch. 3 - Table 3.2.1 gives the inertia IO for a point mass...Ch. 3 - A motor supplies a moment M to the pulley of...Ch. 3 - Figure P3.9 shows an inverted pendulum. Obtain the...Ch. 3 - The two masses shown in Figure P3.10 are released...

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Concept explainers

Videos

The speed and distance traveled of the mass as the functions of time. And the time taken by the mass to reach the platform and ground.

Answer to Problem 3.1P

Explanation of Solution

Want to see more full solutions like this?

Chapter 3 Solutions