Question

A 4.00 kg block is released from rest on a ramp that is inclined at an angle of 60.0∘ below the horizontal. The initial position of the block is a vertical distance of 2.00 m above the bottom of the ramp.

Part A) If the speed of the block is 5.00 m/s when it reaches the bottom of the ramp, what was the work done on it by the friction force?: ANS: -28.4J

Part B) If the angle of the ramp is changed but the block is released from a point that is still 2.00 m above the base of the ramp, both the magnitude of the friction force and the distance along the ramp that the block travels change. If the angle of the incline is changed to 50.0∘, does the magnitude of the work done by the friction force increase or decrease compared to the value calculated in part A?: ANS: The magnitude of the work done by friction increases as θ decreases.

Part C) How much work is done by friction when the ramp angle is 50.0???

Answer 1

apply   work  energy  theorem energy  

Work_by all forces    =change  in  KE

W_gravity + W_friction   = 1/2* m* ( vf^2-vi^2)

mgh + W_friction = 1/2* m*vf^2

4*9.8*2 + W_friction = 0.5*4*5^2

W_friction = 0.5*4*5^2 -4*9.8*2 = -28.4 J answer

work done by friction = f *d = μ *mg*cosθ *d so

if angle θ decreases ,cosθ = increases so work done  increases  

PART C

we need to find out what is μ and d

W_friction =μ *mg*cosθ *d =μ *mg*d *cos60 = 28.4

μ *mg*d = 28.4 /cos60 = 28.4/0.5 = 56.8

so now work done by friction when θ = 50

W_friction = μ *mg*d *cos50 =56.8*cos50 = -36.5103 J answer

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