Question

You drag a crate across a floor with a rope. The force applied is 750 N and the angle of the rope is 25.0° above the horizontalHow much work do you do on the crate when moving it 26.0 m?If you complete the job in 6.00 s, what power is developed?

Answer 1

a.

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

`W=F\cos\theta\cdot d`

where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

`\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}`

Therefore the work done is 17673 J.

b)

The power is given by:

`P=(W)/(t)`

the problem states that the time it takes is 6 s. Then:

`\begin{gathered} P=(17673)/(6) \\ P=2945.5 \end{gathered}`

Therefore the power is 2945.5 W