A small block of mass 1 kg is released from rest at the top of a rough track. The track is a circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure below, is 150 J. (Take the acceleration due to gravity, g = 10 ms–2)
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The magnitude of the normal reaction that acts on the block at the point Q is
= 1 × 10 × 20 – 150 = 50
v = 10 m/s
N – 10 cos 60° =
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The speed of the block when it reaches the point Q is
= 1 × 10 × 20 – 150 = 50
v = 10 m/s
N – 10 cos 60° =
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