Question

To push a 21 kg crate up a frictionless incline, angled at 40° to the horizontal, a worker exerts a force of 224.9 N, parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by the worker's applied force?

Answer 1

Work done, W = 337.35 Joules

Step-by-step explanation:

It is given that,

Mass of the crate, m = 21 kg

Force exerted by the worker on the crate, F = 224.9 N

The crate is inclined at an angle of 40 degrees.

Distance covered by the slide, d = 1.5 m

Let W is the work is done on the crate by the worker's applied force. it is equal to the product of force and distance. It is given by :

`W=F* d`

`W=224.9\ N* 1.5\ m`

W = 337.35 Joules

So, the work is done on the crate by the worker's applied force is 337.35 Joules. Hence, this is the required solution.