A crate of mass 50 kg slides down a 30 degree incline. The crate\'s acceleration

Question

A crate of mass 50 kg slides down a 30 degree incline. The crate's acceleration is 2.0 m/s^2, and the incline is 10 m long, (a) What is the kinetic energy of the crate as it reaches the bottom of the incline? (b) How much work is spent in overcoming friction? (c) What is the magnitude of the frictional force that acts on the crate as it slides down the incline? (d) What is the coefficient of kinetic friction between the crate and the incline? At the base of the incline there is a horizontal surface with the same coefficient of kinetic friction - how far will the crate slide before coming to rest?

Explanation / Answer

a)

Kinetic energy of the crate as it reaches the bottom o the incline

KE=(1/2)mV2

Since V=sqrt[2ah]

KE=mah =50*2*10

KE=1000 J

b)

Work done by the gravity is

Wg=mgh =50*9.8*10*sin30

Wg=2450 J

Work done by friction

Wf=KE-Wg =1000-2450

Wf=-1450 J

c)

Since Work done by friction

Wf=-Ffd

=>Ff=1450/10 =145 N

d)

Coefficent of Kinetic friction

uk =Ff/mgCos30 =145/(50*9.8*cos30)

uk=0.342

e)

The distance travelled by crate before coming to rest

d=KE/ukmg =1000/0.342*50*9.8

d=5.97 m

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