A0.50-$kg$ 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.

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  

current flows in a wire