Chapter 4, Problem 4SP

(a)

To determine

Whether the crate will accelerate.

Answer to Problem 4SP

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

Explanation of Solution

Given info: The mass of the crate is $70\text{kg}$ , 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 $70\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to get $F_g$.

$\begin{array}{c}{F}_g=\left(70\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =686\text{N}\end{array}$

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 $686\text{N}$ for $F_g$ and $500\text{N}$ for $F_{\text{upward}}$ in the above equation to get $F_{\text{net}}$.

$\begin{array}{c}{F}_{\text{net}}=686\text{N}-500\text{N}\\ =186\text{N}\end{array}$

The net force is $186\text{N}$ 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 since there is a net force acting in the downward direction.

(b)

To determine

The magnitude and direction of acceleration.

Answer to Problem 4SP

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

Explanation of Solution

Given info: The mass of the crate is $70\text{kg}$.

Write the expression for the acceleration of the crate.

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

Substitute $186\text{N}$ for $F_{\text{net}}$ and $70\text{kg}$ for $m$ in the above equation to get $a$.

$\begin{array}{c}a=\frac{186\text{N}}{70\text{kg}}\\ =2.66\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the magnitude and direction of acceleration is $2.66\text{m}/{\text{s}}^{\text{2}}$.

(c)

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  $1.03\text{s}$.

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 $v_0$ , $2.66\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $1.4\text{m}$ for $d$ in the above equation to get $t$.

$\begin{array}{c}1.4\text{m}=\left(0\text{m}/\text{s}\right)t+\frac{1}{2}\left(2.66\text{m}/{\text{s}}^{\text{2}}\right)t^2\\ t=1.0259\text{s}\\ \text{=1}\text{.03}\text{s}\end{array}$

Conclusion:

Thus, the time taken by the crate to reach the floor, if the crate is $1.4\text{m}$ above the floor is  $1.03\text{s}$.

(d)

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 $2.7\text{m}/\text{s}$.

Explanation of Solution

Write the expression for the velocity of the crate.

$v=v_0+at$

Substitute $0\text{m}/\text{s}$ for $v_0$ , $2.66\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $1.03\text{s}$ for $t$ in the above equation to get $v$.

$\begin{array}{c}v=0\text{m}/\text{s}+\left(2.66\text{m}/{\text{s}}^{\text{2}}\right)\left(1.03\text{s}\right)\\ =2.7\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of the crate when it hits the floor is $2.7\text{m}/\text{s}$.