Chapter 4, Problem 4SP

A 60-kg crate is lowered from a loading dock to the floor using a rope passing over a fixed support. The rope exerts a constant upward force on the crate of 500 N.

1. a. Will the crate accelerate? Explain.
2. b. What are the magnitude and direction of the acceleration of the crate?
3. c. How long will it take for the crate to reach the floor if the height of the loading dock is 1.4 m above the floor?
4. d. How fast is the crate traveling when it hits the floor?

To determine

Whether the crate will accelerate.

Answer to Problem 4SP

The crate will accelerate downward since there is a net force acting in the downward direction.

Explanation of Solution

Given info: The mass of the crate is $60\text{kg}$ , upward force is $500\text{N}$.

Write the expression for the gravitational force.

$F_g=mg$

Here,

$F_g$ is the gravitational force

$m$ is the mass of the crate

$g$ is the acceleration due to gravity

Substitute $60\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to get $F_g$.

$\begin{array}{c}{F}_g=\left(60\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =588\text{N}\end{array}$

This force acts in the downward direction.

Write the expression for the net force.

$F_{\text{net}}=F_g-F_{\text{upward}}$

Here,

$F_{\text{net }}$ is the net force acting on the crate

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

The negative sign indicate that upward force is opposite to gravitational force.

Substitute $588\text{N}$ for $F_g$ and $500\text{N}$ for $F_{\text{upward}}$ in the above equation to get $F_{\text{net}}$.

$\begin{array}{c}{F}_{\text{net}}=588\text{N}-500\text{N}\\ =88\text{N}\end{array}$

The net force is $88\text{N}$ and is acting in the direction of gravitational force that is , downward direction.

Therefore according to the newton’s second law there will be a downward acceleration for the crate.

Conclusion:

Thus, the crate will accelerate downwad since there is a net force acting in the downward direction.

To determine

The magnitude and direction of acceleration.

Answer to Problem 4SP

The magnitude and direction of acceleration is $1.47\text{m}/{\text{s}}^{\text{2}}$ downward.

Explanation of Solution

Given info: The mass of the crate is $60\text{kg}$ .

Write the expression for the acceleration of the crate.

$a=\frac{F_{\text{net}}}{m}$

Substitute $88\text{N}$ for $F_{\text{net}}$ and $60\text{kg}$ for $m$ in the above equation to get $a$.

$\begin{array}{c}a=\frac{88\text{N}}{60\text{kg}}\\ =1.47\text{m}/{\text{s}}^{\text{2}}\end{array}$

The direction is downward.

Conclusion:

Thus, the magnitude and direction of acceleration is $1.47\text{m}/{\text{s}}^{\text{2}}$ downward.

To determine

The time taken by the crate to reach the floor, if the crate is $1.4\text{m}$ above the floor.

Answer to Problem 4SP

The time taken by the crate to reach the floor, if the crate is $1.4\text{m}$ above the floor is  $1.38\text{s}$.

Explanation of Solution

Given info: The distance is $1.4\text{m}$.

Assume that the crate is rest initially.

Write the expression for the distance travelled by the block.

$d=v_{0}t+\frac{1}{2}a{t}^2$

Here,

$d$ is the distance

$a$ is the acceleration

$t$ is the time

Substitute $0\text{m}/\text{s}$ for $v_0$ , $1.47\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $1.4\text{m}$ for $d$ in the above equation to get $t$.

$\begin{array}{c}1.4\text{m}=\left(0\text{m}/\text{s}\right)t+\frac{1}{2}\left(1.47\text{m}/{\text{s}}^{\text{2}}\right)t^2\\ t=1.38\text{s}\end{array}$

Conclusion:

Thus, the time taken by the crate to reach the floor, if the crate is $1.4\text{m}$ above the floor is  $1.38\text{s}$.

To determine

The velocity of the crate when it hits the floor.

Answer to Problem 4SP

The velocity of the crate when it hits the floor is $2.03\text{m}/\text{s}$.

Explanation of Solution

Write the expression for the velocity of the crate.

$v=v_0+at$

Substitute $0\text{m}/\text{s}$ for $v_0$ , $1.47\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $1.38\text{s}$ for $t$ in the above equation to get $v$.

$\begin{array}{c}v=0\text{m}/\text{s}+\left(1.47\text{m}/{\text{s}}^{\text{2}}\right)\left(1.38\text{s}\right)\\ =2.03\text{s}\end{array}$

Conclusion:

Thus, the velocity of the crate when it hits the floor is $2.03\text{m}/\text{s}$.