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

Question

4 months ago

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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 μk. 1)What is the macroscopic work done on the block by friction during this process? mgH

Physics

1 answer:

kicyunya [3.2K] · Answer 1 · 2026-02-22T19:10:41+03:00

Answer:

$\displaystyle W=-m.g.H$

Explanation:

Transformation of Energy

Also known as energy conversion, this refers to the process in which energy shifts from one type to another. In this context, three energy forms are involved. When the object is stationary at the ramp's peak, it possesses gravitational potential energy, calculated as

$U=m.g.H$

As the object descends the frictionless ramp, it converts all its potential energy into kinetic energy, represented as

$\displaystyle K=\frac{m.v^2}{2}$

Thus,

$K=m.g.H$

Ultimately, when the object encounters a rough surface, all energy converts to thermal energy. The work performed by the friction force corresponds to the alteration in kinetic energy, as all velocity is lost in this process:

$\displaystyle W=\Delta E=K_f-K=0-K=-\frac{m.v^2}{2}$

Given the kinetic energy equals the initial potential energy:

$\boxed{\displaystyle W=-m.g.H}$

The negative sign indicates that the work acted against the direction of movement, meaning the force and displacement are 180° apart.

This outcome is independent of the distance D needed to halt the block or the kinetic friction coefficient.