Q12E - University Physics with Modern Physics

Question

A crate with mass 32.5 kg initially at rest on a warehouse floor is acted on by a net horizontal force of 14.0 N. (a) What acceleration is produced? (b) How far does the crate travel in 10.0 s? (c) What is its speed at the end of 10.0 s?

Step-by-Step Solution

Answer

(a) The acceleration produced in the crate of 32.5 kg initially at rest when acted on by a net horizontal force of 14.0 N is $0.43\mathrm{m}/{\mathrm{s}}^2$

(b) On application of the force, the crate travels 21.5 m in 10 s.

(c) The final velocity of the crate after 10 s is 4.3 m/s.

Step 1: Given data

The mass of the crate is

$\text{m}=32.5 \text{kg}$

The net horizontal force applied to the crate is

$\text{F}=14 \text{N}$

The initial velocity of the block is

$\text{u}=0\mathrm{m}/\mathrm{s}$

The force is applied for

$\text{t}=10 \text{s}$

Step 2: Laws and equations of motion

The distance traveled $s$ , time of travel $t$ , initial velocity $u$  , and acceleration $a$  are related as 

$\mathit{S}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

The second law of motion relates the force $f$ , mass data-custom-editor="chemistry" $\mathrm{m}$  , and acceleration $a$  as

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

The final velocity $\mathrm{v}$ ,  time of travel $t$ , initial velocity  $u$ , and acceleration $a$  are related as

$\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(3\right)$

Step 3: Acceleration of the crate

Let the acceleration of the block be $a$ . From equation (2),

$\begin{array}{rcl}\mathrm{a}& =& \frac{\mathrm{F}}{\mathrm{m}}\\ & =& \frac{14 \text{N}}{32.5 \text{kg}}\\ & =& 0.43 \left(1 {\text{kg}}^{\text{ - 1}}\right)·\left(1\mathrm{N}\times \frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)\end{array}$

Thus, the acceleration of the crate is $0.43\mathrm{m}/{\mathrm{s}}^2$.

Step 4: Distance traveled by the crate

From equation (1), the distance traveled by the crate is

$\begin{array}{rcl}\mathrm{S}& =& \text{ut}+\frac{\text{1}}{\text{2}}{\text{at}}^{\text{2}}\\ & & \text{=}0+\frac{1}{2}\times 0.43\mathrm{m}/{\mathrm{s}}^2\times 100 {\text{s}}^2\\ & & \text{=21.5m}\end{array}$

Thus, the distance traveled by the crate is $21.5\mathrm{m}$

Step 5: Final velocity of the crate

From equation (3), the final velocity of the crate is

$\begin{array}{rcl}\mathrm{v}& =& \text{u}+\text{at}\\ & =& 0+0.43 \mathrm{m}/{\mathrm{s}}^2\times 10 \text{s}\\ & =& 4.3\mathrm{m}/\mathrm{s}\end{array}$

Thus, the final velocity is $4.3\mathrm{m}/\mathrm{s}$