A 50 kg crate slides down a 5.0 m loading ramp that is inclined at an angle of 25 to the horizontal. A worker pushes on the crate parallel to the surface of the ramp so that the crate slides down with a constant velocity. If the coefficient of kinetic friction between the crate and the ramp is 0.33, how much work is done by (a) the worker

The magnitude of the work done by the worker is 303 J.

Explanation:

The work done by the worker can be found as follows:

Where:

F: is the force applied by the worker

d: is the displacement = 5.0 m

α: is the angle between the force applied and the displacement = 180°

We need to find the force applied by the worker:

Taking as positive the movement direction of the crate we have:

Where:              

m: is the crate's mass

a: is the acceleration = 0 (It is moving at constant speed)

F: is the force applied by the worker

Pₓ: is the weight in the horizontal direction    

: is the frictional force  

Hence, the force applied by the worker is:            

Then, the work done by the worker is:                        

Therefore, the magnitude of the work done by the worker is 303 J.

I hope it helps you!                                            

laticia draws the diagram below to show earth’s magnetic field.

answer

the arrows to the right of earth should point up.