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A 5 kg block is set into motion up an inclined plane with an initial speed of 8 m/s. If...
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A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s. The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30 degrees.
Let the coefficient of friction be C. The frictional force that causes the block to come to a rest after traveling 3 m is equal to C*N where N is the normal force. The weight of the block is 5*9.8 = 49 N. This force can be divided into two components one parallel to the incline and one perpendicular to the incline. The component perpendicular to the incline is the normal force equal to 49*cos 30 = 49*sqrt 3/2
The component parallel to the incline acting downwards is 24.5 N. The resistive force to the block is 24.5 + C*24.5*sqrt 3. The deceleration due to this is 4.9(1 + C*sqrt 3)
8^2 = 0^2 = -2*3*4.9*(1 + C*sqrt 3)
=> C = 0.679
The coefficient of friction is 0.679.
Posted by justaguide on March 25, 2012 at 10:20 PM (Answer #1)
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