A mover's dolly is used to pull a 115 kg refrigerator up a ramp into a house. The ramp is 2.10 m long and rises 0.850 m. The
mover exerts a force of 496 N up the ramp. How much work is
spent overcoming friction?

The amount of work  spent in overcoming friction is 83.65 J.

From the question,

The amount of work spent in overcoming friction = work done by the mover-potential energy of the refrigerator.

This is expressed mathematically as,

W' = Fd-mgh Equation 1

Where W' = work spent against friction, F = force applied, d = distance, m = mass of the refrigerator, g = acceleration due to gravity, h = height.

From the question,

Given: F = 496 N, d = 2.10 m, m = 115 kg, h = 0.85 m

Constant: g = 9.8 m/s².

Substitute these values into equation 1

W' = 496(2.1)-(115×9.8×0.85)

W' = 1041.6-957.95

W' = 83.65 J.

Hence, the amount of work  spent in overcoming friction is 83.65 J

Learn more about friction here:

2b2t

explanation:

2b2t

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray