A 4.6 kg block is at the top of a 2.50 m long incline that is at an angle of 37o

Question

A 4.6 kg block is at the top of a 2.50 m long incline that is at an angle of 37o above the horizontal. The incline is frictionless. At the bottom of the incline a horizontal surface has a coefficient of friction given by 0.33 and is 2.10 m long. Note: For full credit use work-energy principles to compute the following: (a) What is the speed of the block at the bottom of the incline? (b) What is the speed of the block after it reaches the end of the horizontal frictioned surface? include force body diagram.

Explanation / Answer

Here ,

m = 4.6 Kg

L = 2.50 m

theta = 37 degree

uk = 0.33

a) let the speed of the block is v

Using work energy theorum

0.50 * m * v^2 = m * g * L * sin(theta)

0.50 * v^2 = 9.8 * 2.50 * sin(37 degree)

solving for v

v = 5.43 m/s

the speed is 5.43 m/s

v) let the speed is v

Using work energy theorum

0.50 * m * (v^2 - 5.43^2) = - 0.33 * m * 9.8 * 2.1

0.50* (v^2 - 5.43^2) = - 0.33 * 9.8 * 2.1

solving for v

v = 4 m/s

the speed of the block is 4 m/s

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