What is the work done on the block by the worker and weight forces?

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SUMMARY

A 36 kg block of ice slides down a frictionless incline measuring 1.2 m in length and 0.48 m in height. The worker exerts a force of 141.2629 N parallel to the incline to maintain constant speed. The work done by the worker's force is -141.2629 J, while the work done by the weight of the block is 141.2629 J. The work is calculated using the formula W = F · d, where F is the force and d is the displacement along the direction of the force.

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A 36 kg block of ice slides down a frictionless incline 1.2 m long and 0.48 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed.
(a) Find the magnitude of the force exerted by the worker.
(b) Find the work done on the block by the workers force.
(c) Find the work done on the block by the weight of the block.

I found the answer to (a) by sin(arcsin(.48/1.2)) * 9.81 * 36 kg which is 141.2629.
I don't know what to do for (b) and (c). My first guess would be that (b) is -141.2629 and (c) is 141.2629. But I am not sure.
 
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Work is equal to the change in kinetic energy. If I push on the wall, am I doing work on the wall?


By the way, a) is correcct, but I don't think it's absolutely necessary to take the sin of the arcsin.
 
Work done BY a Force is F dot displacement .
You know F_by_worker is up-the-ramp, so does neg. W,
but how far does the block travel along that direction?
[ hint: not exactly 1 meter ! ]

(mg acts along the -.48 m Delta_z, so does +'ve W)
 

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