A 0.50-$kg$ block slides along a small track with elevated ends and a flat central part. The flat part has a length L = 1.00 $m$
. The curved portions of the track are frictionless, but for the flat part the coefficient of kinetic friction is 0.144. The block is released from rest from a height h = 70 $cm$ on the left curved portion of the track. Calculate the maximum height reached by the block on the right curved portion of the track.
1 answer:
Answer:
y=55.6 cm
Explanation:
Given that
m=0.5 kg
L= 1 m
h= 70 cm = 0.7 m
coefficient of kinetic friction( μ ) = 0.144
Lets take maximum of height achieved by block on the right side
The friction force on the flat part
fr= μ m g
Initial velocity at the top left is zero and the final speed of the bock at right top will be zero .
we know that work done by all force = Change in kinetic energy
Here initial and final speed is zero so the change in kinetic energy is zero.
m g h - fr . L - m g y=0
fr= μ m g
m g h - μ m g .L - m g y=0
h - μ.L - y = 0
By putting the values
0.7 - 0.144 x 1 - y = 0
y=0.556 m
y=55.6 cm
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