Chapter 6, Problem 1SP

(a)

To determine

The work done by the $6\text{N}$ force.

Answer to Problem 1SP

The work done by the $6\text{N}$ force is $10.2\text{J}$.

Explanation of Solution

Given info: The force is $6\text{N}$ and the distance moved is $1.7\text{m}$.

Write the expression for the work done by a horizontally applied force.

$W=Fd$

Here,

$W$ is the work done by the force

$F$ is the applied force

$d$ is the displacement of the object

Substitute $6\text{N}$ for $F$ and $1.7\text{m}$ for $d$ to find the work done by the $6\text{N}$ force.

$\begin{array}{c}W=\left(6\text{N}\right)\left(1.7\text{m}\right)\\ =10.2\text{J}\end{array}$

Conclusion:

Therefore, the work done by the $6\text{N}$ force is $10.2\text{J}$.

(b)

To determine

The work done by the net force acting on the block.

Answer to Problem 1SP

The work done by the net force acting on the block is $6.8\text{J}$.

Explanation of Solution

Given info: The applied force is $6\text{N}$, the frictional force is $2\text{N}$ and the distance moved by the block is $1.7\text{m}$.

Write the expression for the work done by a horizontally acting forces.

$W=F_{\text{total}}d$ (1)

Here,

$F_{\text{total}}$ is the total forces acting horizontally to the body

Write the expression for the total force.

$F_{\text{total}}=F_{\text{applied}}-F_{\text{friction}}$ (2)

Here,

$F_{\text{applied}}$ is the applied force

$F_{\text{friction}}$ is the frictional force

Use equation (2) in (1).

$W=\left(F_{\text{applied}}-F_{\text{friction}}\right)d$

Substitute $6\text{N}$ for $F_{\text{applied}}$, $2\text{N}$ for $F_{\text{friction}}$ and $1.7\text{m}$ for $d$ to find the work done $W$.

$\begin{array}{c}W=\left(6\text{N}-2\text{N}\right)\left(1.7\text{m}\right)\\ =6.8\text{J}\end{array}$

Conclusion:

Therefore, the work done by the net force acting on the block is $6.8\text{J}$.

(c)

To determine

Which among the two values of work done should be used find the increase in kinetic energy of the block.

Answer to Problem 1SP

$6.8\text{J}$, which is the work done by the net force acting on the block should be used to find the increase in kinetic energy of the block.

Explanation of Solution

Given info: The work done by the applied force is $10.2\text{J}$ and the work done by the net force is $6.8\text{J}$.

According to the work energy theorem, the change in kinetic energy is equal to the net work done on the object. The increase in kinetic energy can be found out only using the net work done on the object, not the work done by any of the applied force.

Thus, in the case of the block the value of work done $6.8\text{J}$, which is the work done by the net force acting on the block should be used to find the increase in kinetic energy of the block.

Conclusion:

Therefore, the work done by the net force, $6.8\text{J}$ acting on the block should be used to find the increase in kinetic energy of the block.

(d)

To determine

What happens to the energy added to the system via the applied force of $6\text{N}$ and whether it can all be accounted for or not.

Answer to Problem 1SP

$3/5$ of the energy (work done) by the applied force goes into changing kinetic  energy and the remaining is converted to heat energy and the all this energy can be accounted for.

Explanation of Solution

The total energy added to the system via applied force is of $10.2\text{J}$ and out of which dissipation due to friction causes the net energy to be $6.8\text{J}$ which is the $3/5$ part of the total energy added to the system via applied force. The rest of the energy is dissipated as heat due to the friction. Moreover, all this energy can be accounted for since the energy is always conserved for all kinds considered.

Conclusion:

Thus, $3/5$ of the energy (work done) by the applied force goes into changing kinetic  energy and the remaining is converted to heat energy and the all this energy can be accounted for.

(e)

To determine

The kinetic energy and velocity of the block at the end of the $1.7\text{m}$ motion.

Answer to Problem 1SP

At the end of the $1.7\text{m}$ motion, the kinetic energy of the block is $6.8\text{J}$ and the velocity is $5.98\text{m/s}$.

Explanation of Solution

Given info: The mass of the block is $0.38\text{kg}$ and the net work done is $6.8\text{J}$.

According to the work energy theorem, the net work done is equal to the change in kinetic energy. Since, the block starts from rest, its initial kinetic energy is zero and hence the change in kinetic energy will be equal to the final kinetic energy.

Thus, the kinetic energy of the block is equal to the net work done, which is equal to $6.8\text{J}$.

$KE=6.8\text{J}$

Write the expression for kinetic energy.

$KE=\frac{1}{2}m{v}^2$

Here,

$m$ is the mass

$v$ is the velocity

Solve for $v$.

$v=\sqrt{\frac{2\left(KE\right)}{m}}$

Substitute $6.8\text{J}$ for $KE$ and $0.38\text{kg}$ for $m$ to find the velocity $v$ .

$\begin{array}{c}v=\sqrt{\frac{2\left(6.8\text{J}\right)}{0.38\text{kg}}}\\ =5.98\text{m/s}\end{array}$

Conclusion:

Therefore, at the end of the $1.7\text{m}$ motion, the kinetic energy of the block is $6.8\text{J}$ and the velocity is $5.98\text{m/s}$.