Chapter 6, Problem 2SP

(a)

To determine

The acceleration of the crate.

Answer to Problem 2SP

The acceleration of the crate is $0.5{\text{ m/s}}^2$.

Explanation of Solution

Given Info: The mass of the crate is $100\text{ kg}$ and the net force on the crate is $50\text{ N}$.

Write the equation for the net force acting on a body given by the Newton’s second law.

$F=ma$

Here,

$F$ is the net force on the crate

$m$ is the mass of the crate

$a$ is the acceleration of the crate

Rewrite the above equation for $a$.

$a=\frac{F}{m}$

Substitute $50\text{ N}$ for $F$ and $100\text{ kg}$ for $m$ in the above equation to find $a$.

$\begin{array}{c}a=\frac{50\text{ N}}{100\text{ kg}}\\ =0.5{\text{ m/s}}^2\end{array}$

Conclusion:

Thus the acceleration of the crate is $0.5{\text{ m/s}}^2$.

(b)

To determine

The distance travelled by the crate in $4\text{ s}$ if it was started from rest.

Answer to Problem 2SP

The distance travelled by the crate in $4\text{ s}$ if it was started from rest is $4\text{ m}$.

Explanation of Solution

Write the equation for the distance displacement.

$s=ut+\frac{1}{2}a{t}^2$

Here,

$s$ is the displacement of the crate

$u$ is the initial velocity of the crate

$t$ is the time taken for the displacement

Substitute $0\text{ m/s}$ for $u$, $0.5{\text{ m/s}}^2$ for $a$ and $4\text{ s}$ for $t$ in the above equation to find $s$.

$\begin{array}{c}s=0+\frac{1}{2}\left(0.5{\text{ m/s}}^2\right){\left(4\text{ s}\right)}^2\\ =4\text{ m}\end{array}$

Conclusion:

Thus the distance travelled by the crate in $4\text{ s}$ if it was started from rest is $4\text{ m}$.

(c)

To determine

The work done by the $50\text{ N}$ force.

Answer to Problem 2SP

The work done by the $50\text{ N}$ force is $200\text{ J}$.

Explanation of Solution

Write the equation for the work done.

$W=Fd$

Here,

$W$ is the work done

$F$ is the applied force

$d$ is the distance moved

Substitute $50\text{ N}$ for $F$ and $4\text{ m}$ for $d$ in the above equation to find $W$.

$\begin{array}{c}W=\left(50\text{ N}\right)\left(4\text{ m}\right)\\ =200\text{ J}\end{array}$

Conclusion:

Thus the work done by the $50\text{ N}$ force is $200\text{ J}$.

(d)

To determine

The velocity of the crate at the end of $4\text{ s}$.

Answer to Problem 2SP

The velocity of the crate at the end of $4\text{ s}$ is $2\text{ m/s}$.

Explanation of Solution

Write the equation for the velocity of the crate.

$v=u+at$

Here,

$v$ is the final velocity of the crate

Substitute $0\text{ m/s}$ for $u$, $0.5{\text{ m/s}}^2$ for $a$ and $4\text{ s}$ for $t$ in the above equation to find $v$.

$\begin{array}{c}v=0\text{ m/s}+\left(0.5{\text{ m/s}}^2\right)\left(4\text{ s}\right)\\ =2\text{ m/s}\end{array}$

Conclusion:

Thus the velocity of the crate at the end of $4\text{ s}$ is $2\text{ m/s}$.

(e)

To determine

The kinetic energy of the crate at the end of $4\text{ s}$ and how does it compare to the work computed in part (c).

Answer to Problem 2SP

The kinetic energy of the crate at the end of $4\text{ s}$ is $200\text{ J}$ and it is equal to the work computed in part (c).

Explanation of Solution

Given Info: The mass of the crate is $100\text{ kg}$.

Write the equation for kinetic energy.

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

Here,

$KE$ is the kinetic energy

Substitute $100\text{ kg}$ for $m$ and $2\text{ m/s}$ for $v$ in the above equation to find $K.E$.

$\begin{array}{c}\text{K}\text{.E}=\frac{1}{2}\left(100\text{ kg}\right){\left(2\text{ m/s}\right)}^2\\ =200\text{ J}\end{array}$

The work computed in part (c) is also equal to $200\text{ J}$.

Conclusion:

Thus the kinetic energy of the crate at the end of $4\text{ s}$ is $200\text{ J}$ and it is equal to the work computed in part (c).