When the block is launched, it gets through the frictional part, rises up the ra

Question

When the block is launched, it gets through the frictional part, rises up the ramp to point D, a height h above the straight part of the ramp. It then slides back down. On its return, it gets to point B and stops. The distances are shown on the figure, and the angle of the ramp is ? as shown.

Given the following values for the parameters, can you calculate the value of the coefficient of friction, ? ? If you can, put the value in the box and explain how you got it. If not, explain why not and what information would allow you to calculate it.

m = 75 g

k = 15 N/m

d1 = 10 cm

d2 = 10 cm

d3 = 8 cm

L = 25 cm

? = 20o

If you need the earth's gravitational field, use the approximate value g = 10 N/kg.
? =_______

Explanation / Answer

Normal force acting on the block, N = m*g

while the block moves left the workdone by friction = -m*g*h

so, when the block moves right also the workdone by friction = -m*g*h

total workdone by friction = -2*m*g*h

we know, Workdone by friction = loss of mechincal energy

fk*L*cos(180) = 0 - 2*m*g*h

-N*mue_k*L = -2*m*g*h

m*g*mue_k*L = 2*m*g*h

mue_k = 2*h/L

= 2*d3*tan(theta)/L

= 2*0.08*tan(20)/0.25

= 0.233 <<<<<<<--------------Answer

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