Question

An 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of 30o with the horizontal. The force pushing the crate is parallel to the slope. If the speed of the crate increases at a rate of 1.5 m/s2, find the work done by the force. Group of answer choices 260 J 200 J 1 J 200 kJ 140 J

Answer 1

The work done by the force is 140 J.

Calculation:

It is given that, an 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of $30^\circ$ with the horizontal. The force pushing the crate is parallel to the slope and the acceleration is 1.5 $\text{ m/s}^2$.

It is required to find the work done by the force.

The free-body diagram is shown below,

Here, W=80 N, a=-1.5 $\text{ m/s}^2$

We can write the net force equation as,

$\begin{gathered}F-W \sin 30^{\circ}=m a \\F-m g \sin 30^{\circ}=\left(\frac{W}{g}\right) a\end{gathered}$

Here W is the weight of the object, m is the mass of the body, and F is the applied force.

Thus,

$\begin{aligned}F-(80 \mathrm{~N}) \sin 30^{\circ} &=\left(\frac{80 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}\right)\left(-1.5 \mathrm{~m} / \mathrm{s}^{2}\right) \\F &=27.755 \mathrm{~N}\end{aligned}$

It is known that the work done by the force is calculated as the multiplication of the applied force and the displacement.

Thus the work one is,

$\begin{aligned}W &=(27.755 \mathrm{~N})(5 \mathrm{~m}) \\& \approx 140 \mathrm{~J}\end{aligned}$

Thus the last option is the correct one.

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