Question

A block of mass m, initially held at rest on a frictionless ramp a vertical distance H above the floor, slides down the ramp and onto a floor where friction causes it to stop a distance D from the bottom of the ramp. The coefficient of kinetic friction between the box and the floor is ?k.

What is the macroscopic work done on the block by all forces during this process?

Answer 1

Step-by-step explanation:

Check attachment for solution

Answer 2

Answer and Explanation:

Total macroscopic workdone = Δ K.E

Where K.E = Kinetic Energy

Let V₁ = Initial speed of the block and

V₂ = Final speed of the block

Δ K.E = 1/2 MV₂² - 1/2 MV₁²

Δ K.E = 1/2M(V₂-V₁)²

Since the block was initially at rest, V₁ = 0

Since the block eventually stopped due to friction, V₂ = 0

Δ K.E = 1/2M * 0

Δ K.E = 0 Joules

N.B: The workdone by gravity is Mgh