The upper half of an inclined plane of inclination theta is perfectly

For first half acceleration = g sin theta Therefore, velocity after travelling ball distance v^(2) = 2(g sin theta)l For second half, acceleration = g(sin theta - mu(k) cos theta) so 0^(2) = v^(2) = 2g (sin phi - mu(k) cos phi)l Solving (i) and (ii) we get mu(k) = 2 tan theta

The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough a block starting from rest at the top of the plane will again come

the 15] (3) ma g+a g-a 13. The upper half of an inclined plane of inclination is perfectly smooth while lower half is rough. A block starting from rest the of the

SOLVED: The upper portion of an inclined plane of inclination alpha is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to

The upper half of an inclined plane of inclination is perfectly smooth while lower half is rough. A block

If A body is released from the top of an inclined plane of inclination theta. It reaches the bottom with velocity V. If keeping the length same the angle of the inclination

CU 10. W TOUN (B) 500N (l) 349 N The upper half of an inclined plane of inclination is perfectly smooth while the lower half is rough. A body starting from the

Solved] the upper half of an inclined plane with inclination (theta) is perfectly smooth while the lower

The upper half of an inclined plane of inclination andtheta; is perfectly smooth while lower half is rough. A blockstarting from rest at the top of the plane will again come to