Chapter 6, Problem 1SP

(a)

To determine

The work done by the $5\text{ N}$ force.

Answer to Problem 1SP

The work done by the $5\text{ N}$ force is $7.5\text{ J}$.

Explanation of Solution

Given Info: The distance moved by the block is $1.5\text{ m}$.

Write the equation for the work done.

$W=Fd$ (1)

Here,

$W$ is the work done

$F$ is the applied force

$d$ is the distance moved

Substitute $5\text{ N}$ for $F$ and $1.5\text{ m}$ for $d$ in equation (1) to find $W$.

$\begin{array}{c}W=\left(5\text{ N}\right)\left(1.5\text{ m}\right)\\ =7.5\text{ J}\end{array}$

Conclusion:

Thus the work done by the $5\text{ N}$ force is $7.5\text{ J}$.

(b)

To determine

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

Answer to Problem 1SP

The work done by the net force acting upon the block is $4.5\text{ J}$.

Explanation of Solution

Given Info: There are two horizontal forces acting on the block; $5\text{ N}$ force pulling the block and $2\text{ N}$ frictional force opposing the motion.

Write the equation for the net force acting on the block.

$\begin{array}{c}F=5\text{ N}-2\text{ N}\\ =3\text{ N}\end{array}$

The difference is taken since the force are acting opposite to each other.

Substitute $\text{3 N}$ for $F$ and $1.5\text{ m}$ for $d$ in equation (1) to find $W$.

$\begin{array}{c}W=\left(\text{3 N}\right)\left(1.5\text{ m}\right)\\ =4.5\text{ J}\end{array}$

Conclusion:

Thus the work done by the net force acting upon the block is $4.5\text{ J}$.

(c)

To determine

The value which among the two work done should be used to find the increase in kinetic energy of the block.

Answer to Problem 1SP

The work done by the net force acting upon the block found in part (b) should be used to find the increase in kinetic energy of the block.

Explanation of Solution

Kinetic energy of an object is the energy of the object associated with its motion. The kinetic energy is equal to one-half the mass of the object times the square of its speed. Work is the force times displacement of an object.

Doing work on an object increases its energy. Since the work involves the transfer of energy, the amount of kinetic energy gained by the object should be equal to the amount of net work done on it. If the object was initially at rest, the work done on the object become equal to its kinetic energy.

The work done by the net force is the net work done on the block. The increase in kinetic energy is equal to the net work done on the object.

Conclusion:

Thus the work done by the net force acting upon the block found in part (b) 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 work done by the $5\text{ N}$ force and can it all be accounted for.

Answer to Problem 1SP

The $\left(3/5\right)$ portion of the energy is used to increase the kinetic energy of the block and the remaining is converted into thermal energy. All of the energy can be accounted for.

Explanation of Solution

The principle of conservation of energy states that energy can neither be created nor be destroyed. It can only be converted from one form to another so that total energy remains constant.

The increase in kinetic energy of the block is $4.5\text{ J}$ so that $4.5\text{ J}$ out of the $7.5\text{ J}$ energy produced by the $5\text{ N}$ force is converted into the kinetic energy of the block. This implies $\left(3/5\right)$ portion of the energy is used to increase the kinetic energy of the block.

The frictional force is opposing the motion of the block. Remaining energy is used to oppose this frictional force and it is thus converted into thermal energy. All of the energy produced by the $5\text{ N}$ force can be accounted for since according to the energy conservation principle energy can never be lost.

Conclusion:

Thus the $\left(3/5\right)$ portion of the energy is used to increase the kinetic energy of the block and the remaining is converted into thermal energy. All of the energy can be accounted for.

(e)

To determine

The kinetic energy and velocity of the block at the end of $1.5\text{ m}$ motion if it was started from rest.

Answer to Problem 1SP

The kinetic energy of the block is $4.5\text{ J}$ and its velocity is $6\text{ m/s}$.

Explanation of Solution

Given Info: The mass of the block is $0.25\text{ kg}$.

If the object was initially at rest, the net work done on the object become equal to its kinetic energy.

$W=KE$

Here,

$W$ is the total work done on the block

$KE$ is the kinetic energy of the block

Write the equation for kinetic energy.

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

Here,

$m$ is the mass of the block

$v$ is the magnitude of the velocity of the block

Rewrite the above equation for $v$.

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

Substitute $4.5\text{ J}$ for $KE$ and $0.25\text{ kg}$ for $m$ in the above equation to find $v$.

$\begin{array}{c}v=\sqrt{\frac{2\times 4.5\text{ J}}{0.25\text{ kg}}}\\ =6\text{ m/s}\end{array}$

Conclusion:

Thus the kinetic energy of the block is $4.5\text{ J}$ and its velocity is $6\text{ m/s}$.