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

laws of motion class 11 neet questions - NAWENDU CLASSES

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

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 upper half of an inclined plane of inclination theta is perfectly smooth while the lower half rough. A block starting from rest the of the plane will again come to rest

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

3. 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 plane will again come to rest

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 θ is perfectly smooth while the lower half is