Chapter 6, Problem 2SP

As described in example box 6.2, a 120-kg crate is accelerated by a net force of 96 N applied for 4 s.

1. a. What is the acceleration of the crate from Newton’s second law?
2. b. If it starts from rest, how far does it travel in the time of 4 s? (See section 2.5 in chapter 2.)
3. c. How much work is done by the 96-N net force?
4. d. What is the velocity of the crate at the end of 4 s?
5. e. What is the kinetic energy of the crate at this time? How does this value compare to the work computed in part c?

To determine

The acceleration of the crate from Newton’s second law.

Answer to Problem 2SP

The acceleration of the crate from Newton’s second law is $0.8{\text{m/s}}^{\text{2}}$.

Explanation of Solution

Given info: The mass of the crate is $120\text{kg}$ and the accelerating force is $96\text{N}$.

Write the expression for the force according to Newton’s second law.

$F=ma$

Here,

$F$ is the force

$m$ is the mass

$a$ is the acceleration

Solve for $a$.

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

Substitute $120\text{kg}$ for $m$ and $96\text{N}$ for $F$ to find the acceleration $a$.

$\begin{array}{c}a=\frac{96\text{N}}{120\text{kg}}\\ =0.8{\text{m/s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the acceleration of the crate from Newton’s second law is $0.8{\text{m/s}}^{\text{2}}$.

To determine

The distance travelled by the crate in $4.0\text{s}$.

Answer to Problem 2SP

The distance travelled by the crate in $4.0\text{s}$ is $6.4\text{m}$.

Explanation of Solution

Given info: The acceleration is $0.8{\text{m/s}}^{\text{2}}$ and the time is $4.0\text{s}$.

Write the expression for the displacement from the fundamental equations of motion.

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

Here,

$s$ is the displacement

$u$ is the initial velocity

$a$ is the acceleration

$t$ is the time

Since the crate is accelerated from rest, the initial velocity is zero. Thus the expression for $s$ reduces to,

$s=a{t}^2$

Substitute $0.8{\text{m/s}}^{\text{2}}$ for $s$, $4.0\text{s}$ or $t$ to find the distance travelled by the crate $s$.

$\begin{array}{c}s=\frac{1}{2}\left(0.8{\text{m/s}}^{\text{2}}\right){\left(4.0\text{s}\right)}^2\\ =6.4\text{m}\end{array}$

Conclusion:

Therefore, the distance travelled by the crate in $4.0\text{s}$ is $6.4\text{m}$.

To determine

The work done by the $96\text{N}$ net force.

Answer to Problem 2SP

The work done by the $96\text{N}$ net force is $614.4\text{J}$.

Explanation of Solution

Given info: The force is $96\text{N}$ and the distance travelled is $6.4\text{m}$.

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 $96\text{N}$ for $F$ and $6.4\text{m}$ for $d$ to find the work done by the applied force.

$\begin{array}{c}W=\left(96\text{N}\right)\left(6.4\text{m}\right)\\ =614.4\text{J}\end{array}$

Conclusion:

Therefore, the work done by the $96\text{N}$ net force is $614.4\text{J}$.

To determine

The velocity of the crate by the end of $4.0\text{s}$.

Answer to Problem 2SP

The velocity of the crate by the end of $4.0\text{s}$ is $3.2\text{m/s}$.

Explanation of Solution

Given info: The acceleration is $0.8{\text{m/s}}^{\text{2}}$ and the time is $4.0\text{s}$.

Write the expression for the final velocity from the fundamental equations of motion.

$v=u+at$

Since the initial velocity is zero, the equation for $v$ reduces to,

$v=at$

Substitute $0.8{\text{m/s}}^{\text{2}}$ for $a$ and $4.0\text{s}$ for $t$ to find the velocity of the crate by the end of $4.0\text{s}$.

$\begin{array}{c}v=\left(0.8{\text{m/s}}^{\text{2}}\right)\left(4.0\text{s}\right)\\ =3.2\text{m/s}\end{array}$

Conclusion:

Therefore, the velocity of the crate by the end of $4.0\text{s}$ is $3.2\text{m/s}$.

To determine

The kinetic energy of the crate at the time $4.0\text{s}$ and how does it compare with the work done by the $96\text{N}$ force.

Answer to Problem 2SP

The kinetic energy of the crate at the time $4.0\text{s}$ is $614.4\text{J}$ and it is equal to the amount of work done by the $96\text{N}$ force.

Explanation of Solution

Given info: The mass of the crate is $120\text{kg}$ and the velocity of the crate at the time $4.0\text{s}$ is $3.2\text{m/s}$ and the work done by the $96\text{N}$ force is $614.4\text{J}$.

Write the expression for the kinetic energy of an object.

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

Here,

$KE$ is the kinetic energy

$m$ is the mass of the object

$v$ is the velocity of the object

Substitute $120\text{kg}$ for $m$ and $3.2\text{m/s}$ for $v$ to find the kinetic energy of the crate by the end of the time $4.0\text{s}$.

$\begin{array}{c}KE=\frac{1}{2}\left(120\text{kg}\right){\left(3.2\text{m/s}\right)}^2\\ =614.4\text{J}\end{array}$

The kinetic energy of the crate by the end of the time $4.0\text{s}$, which is accelerated from rest by the force of $96\text{N}$, is equal to the work done by the same force on the crate.

Conclusion:

Therefore, the kinetic energy of the crate at the time $4.0\text{s}$ is $614.4\text{J}$ and it is equal to the amount of work done by the $96\text{N}$ force.