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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?

Respuesta :

Answer:

Work done, W = 337.35 Joules

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 :

[tex]W=F\times d[/tex]

[tex]W=224.9\ N\times 1.5\ m[/tex]    

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.

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