Question

# A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of...

A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introduced at point A, brings the block to rest at point B. If the coefficient of kinetic friction is 0.26, what is the distance between A and B?

Step 1:

Using Energy conservation for frictionless incline

KEi + PEi = KEf + PEf

KEi = 0, since block slides from rest

PEi = m*g*h

KEf = 0.5*m*Vf^2

where, Vf = speed of block at bottom of the incline

PEf = 0, since h = 0 at this point

So,

0 + m*g*h = 0.5*m*Vf^2

Vf = sqrt (2*g*h)

Vf = sqrt (2*9.8*5)

Vf = 9.9 m/sec

Now during frictional surface, Using energy conservation

KEi + PEi + W = KEf + PEf

PEi = PEf = 0, since h = 0

KEf = 0, since block stops after traveling some distance 'd', So at point B, speed is zero

KEi = 0.5*m*Vi^2,

here Vi = 9.9 m/sec = Velocity at the bottom of incline at point A

W = Work-done by frictional force

W = -Ff*d = -uk*N*d = -uk*m*g*d

So,

0.5*m*Vi^2 + 0 - uk*m*g*d = 0 + 0

d = Vi^2/(2*uk*g)

Using known values

d = 9.9^2/(2*0.26*9.8)

d = 19.2 m = distance between A and B

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