FrictionHard
Question
A block lying on a long horizontal conveyor belt moving at a constant velocity receives a velocity 5 m/s relative to the ground in the direction opposite to the direction of motion of the conveyor. After t = 4 sec, the velocity of the block becomes equal to the velocity of the belt. The coefficient of friction between the block and the belt is 0.2. Then the velocity of the conveyor belt is : (g = 10 m/s2)
Options
A.13 m/s
B.−13 m/s
C.3 m/s
D.6 m/s
Solution
Since there is relative motion between block and conveyor, there will be frictional force = μN = 0.2 × m × 10
retardation =
= 2 m/sec2
Now v = u + at
⇒ v = 5 + (−2) 4
= −3 = 3 m/s opposite so the direction of initial velocity of the block.
retardation =
= 2 m/sec2 Now v = u + at
⇒ v = 5 + (−2) 4
= −3 = 3 m/s opposite so the direction of initial velocity of the block.
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