A 10.00 kg block starts from rest at the bottom of a 30.00° inclined plane. The
ID: 1642952 • Letter: A
Question
A 10.00 kg block starts from rest at the bottom of a 30.00° inclined plane. The coefficient of kinetic friction between the block and the plane is k=0.10.
a) If the block is launched with an initial velocity of 3.00 m/s, how much distance will it take for the block’s velocity to drop to zero?
b) Assuming the static friction isn’t enough to keep the block stationary when velocity drops to zero, how much time will it take the block to return to the bottom of the plane?
c) Does it take the same time to reach the peak as it does to return to the bottom from its peak height? Explain your answer
Explanation / Answer
(A) Along the incline,
Fnet = m a
- m g sin30 - uk m g cos30 = m a
a = - 5.75 m/s^2
Applying vf^2 - vi^2 = 2 a d
0^2 - 3^2 = 2(-5.75)(d)
d = 0.78 m
(b) While sliding down,
Fnet = m g sin(30) - uk m g cos30 = m a
a = 9.8 (sin30 - 0.1cos30) = 0.413 m/s^2
d = u t + a t^2 / 2
0.78 =0 + 0.413 t^2 /2
t = 1.94 sec
(c)No time will not be same.
friction force always opposes the motion.
but gravity oppose while going up and support while going down.
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