Problem 6.29 A 900-kg car initially at rest rolls 50 m down a hill inclined at a
ID: 1453979 • Letter: P
Question
Problem 6.29 A 900-kg car initially at rest rolls 50 m down a hill inclined at an angle of 5.0. A 400-N effective friction force opposes its motion. Part A How fast is the car moving at the bottom? Express your answer to two significant figures and include the appropriate units. Part B What distance will it travel on a similar horizontal surface (assume the friction force is the same) at the bottom of the hill? Express your answer to two significant figures and include the appropriate units. Part C If the friction force remains the same will the distance decrease or increase if the car's mass is 1800 kg? Part D Determine the new distance. Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
acceleration=(mgsinx-Friction force)/m
a=(900*9.8*sin5-400)/900
a=0.41 m/s2
v2=u2+2as
v2=0+2(0.41)(50)
v=6.40 m/s
Part B) distance covered on horizontal surface=s
acceleration=-(Friction force)/m
a=-400/900
a=-0.44 m/s2
v=6.40 =initial velocity at bottom of hill
final velocity=0
0=6.42-2(0.44)s
s=46.54 m
Part C
as
acceleration=(-Friction force)/m
a=-(Friction force)/m
so friction force remains same and mass increases so (Friction force)/m decreases so acceleration=-(Friction force)/m increases
so distance covered increases
Part D
acceleration=(mgsinx-Friction force)/m
a=(1800*9.8*sin5-400)/1800
a=0.63 m/s2
v2=u2+2as
v2=0+2(0.63)(50)
v=7.94 m/s
distance covered on horizontal surface=s
acceleration=-(Friction force)/m
a=-400/1800
a=-0.22 m/s2
v=7.94 m/s =initial velocity at bottom of hill
final velocity=0
0=7.942-2(0.22)s
s=143.18m