Chapter 4, Problem 2SP

(a)

To determine

The acceleration of the crate.

Answer to Problem 2SP

The acceleration of the crate is $\text{3}\text{m}/{\text{s}}^{\text{2}}$ in the direction of velocity.

Explanation of Solution

Given info: The horizontal force is $280\text{N}$, initial velocity is $1\text{m}/\text{s}$ ,final velocity is $7\text{m}/\text{s}$ and time for the change in velocity is $2\text{s}$.

Write the equation of motion of the crate.

$v=v_0+at$

Here,

$v$ is the velocity of the crate

$v_0$ is the initial velocity of the crate

$a$ is the acceleration of the crate

$t$ is the time

Rearrange the above equation to get $a$.

$a=\frac{v-v_0}{t}$

Substitute $1\text{m}/\text{s}$ for $v_0$ ,$7\text{m}/\text{s}$ for $v$ and $2\text{s}$ for $t$ in the above equation to get $a$.

$\begin{array}{c}a=\frac{7\text{m}/\text{s}-1\text{m}/\text{s}}{2\text{s}}\\ =3\text{m}/{\text{s}}^{\text{2}}\end{array}$

The direction of the crate is in direction of velocity.

Conclusion:

Thus, the acceleration of the crate is $\text{3}\text{m}/{\text{s}}^{\text{2}}$ in the direction of velocity

(b)

To determine

The net force acting on the crate.

Answer to Problem 2SP

The net force acting on the crate $150\text{N}$ in the direction of the velocity.

Explanation of Solution

Given info: The mass of the crate is $50\text{kg}$.

Write the expression for the net force.

$F_{\text{net}}=ma$

Here,

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

$m$ is the mass of the crate

$a$ is the net acceleration of the crate

Substitute $50\text{kg}$ for $m$ and $\text{3}\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to get $F_{\text{net}}$.

$\begin{array}{c}{F}_{\text{net}}=\left(50\text{kg}\right)\left(3\text{m}/{\text{s}}^{\text{2}}\right)\\ =150\text{N}\end{array}$

According to the second law motion this force will be in the direction of acceleration. This implies that the force is in the direction of velocity.

Conclusion:

Thus, the net force acting on the crate $150\text{N}$ in the direction of the velocity.

(c)

To determine

The magnitude of the frictional force acting on the crate.

Answer to Problem 2SP

The magnitude of the frictional force acting on the crate is $130\text{N}$.

Explanation of Solution

Write the expression for the net horizontal force.

$F_{\text{net}}=F_{\text{constant}}-F_{\text{friction}}$

Here,

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

$F_{\text{constant}}$ is the horizontal force

$F_{\text{friction}}$ is the frictional force

The negative sign indicate that frictional force is opposite to horizontal force

Substitute $280\text{N}$ for $F_{\text{constant}}$ and $150\text{N}$ for $F_{\text{net}}$ in the above equation to get $F_{\text{friction}}$.

$\begin{array}{c}150\text{N}=280\text{N}-F_{\text{friction}}\\ F_{\text{friction}}=280\text{N}-150\text{N}\\ \text{=130}\text{N}\end{array}$

Conclusion:

Thus, the magnitude of the frictional force acting on the crate is $130\text{N}$.

(d)

To determine

The force that must apply to crate by the rope in order for the crate to move with constant velocity.

Answer to Problem 2SP

The force that must apply to crate by the rope in order for the crate to move with constant velocity is $130\text{N}$

Explanation of Solution

Write the expression for the net horizontal force.

$F_{\text{net}}=F_{\text{constant}}-F_{\text{friction}}$

The frictional force is $190\text{N}$. The crate will move with a constant velocity if the acceleration is zero.

Substitute $0\text{N}$ for $F_{\text{net}}$ and $130\text{N}$ for $F_{\text{friction}}$ in the above equation to get $F_{\text{constant}}$.

$\begin{array}{c}0\text{N}=F_{\text{constant}}-130\text{N}\\ F_{\text{constant}}=130\text{N}\end{array}$

Conclusion:

Thus, the force that must apply to crate by the rope in order for the crate to move with constant velocity is $130\text{N}$