A 20-kg crate slides along a horizontal frictionless surface at a constant speed

ID: 2144142 • Letter: A

Question

A 20-kg crate slides along a horizontal frictionless surface at a constant speed of 4.0 m/s.

The crate then slides down a frictionless incline and across a second-lower horizontal

surface as shown in the figure. The lower horizontal surface is not frictionless. Define

zero height as the level of the second horizontal surface.

b. What is the speed of the crate when it arrives at the lower surface?

c. What is the coefficient of kinetic friction required to bring the crate to a stop over

a distance of 5.0 m along the lower surface?

d. What work has been done on the crate to stop it?

first of all figure is not provided here.

let , h be the height of incline,

for calculating the speed at lower surfaces,

we'll conserve the mechanical energy.

mgh+ 1/2 m* u^2 = 1/2*mv^2

v = sqrt(4^2+2*9.8*h)

to stop over a distance of 5.0 m along the lower surface.

let mu be the coefficient of kinetic friction,

deceleration (-ve acceleration) = -mu* g,

using, v^2 = u^2 + 2as,

final vel = 0,

initial vel = sqrt(4^2+2*9.8*h)

a= -mu* g,, s = 5m,

we can find 'mu ' from this.

3. work done in stopping = Friction force * displacement

W = mu*M*g * 5

from here we can find 'W'

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