A brick with mass m= 1.3 kg slides on an inclined plane with kinetic coefficient

Question

A brick with mass m= 1.3 kg slides on an inclined plane with kinetic coefficient of friction m_k=0.2 and angle of inclination theta =25degree. Starting from rest at height h_i=1.8 m, the brick slides towards the bottom where it compresses an ideal spring with spring constant k=12 10^3 N/m. The spring then pushes the brick back up the inclined plane. A sketch of this situation is shown in Fig. 1. You can ignore friction on the horizontal part of the track. Calculate the maximum compression of the spring. Calculate the maximum height that the brick reaches as it moves back up the inclined plane.

Explanation / Answer

While moving down:
distance to be travelled on incline, d = h/sin 25 = 1.8/sin 25 = 4.26 m

frictional force acting = miu*m*g*cos thetha = 0.2*1.3*9.8*cos 25 = 2.31 N
energy lost against friction = f*d = 2.31*4.26 = 9.84 J

initial potential energy = m*g*h = 1.3*9.8*1.8 = 22.93 J

use:
initial potential energy = energy lost against friction + spring potential energy
22.93 = 9.84 +0.5*k*x^2
22.93 = 9.84 +0.5*12*10^3*x^2
x=0.047 m
Maximum compression = 0.047 m = 4.7 cm

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while going up:

lets assume it reached height h
distance to be travelled on incline, d = h/sin 25 = 2.37 h
energy lost against friction = f*d = 2.31*2.37 h = 5.47h

use:
initial spring potential energy = final potential energy + energy lost against friction
0.5*k*x^2 = m*g*h + 5.47h
0.5*12*10^3*(0.047)^2 = 1.3*9.8*h + 5.47h
0.5*12*10^3*(0.047)^2 = 1.3*9.8*h + 5.47h
13.09 = 12.74h + 5.47h
h=0.72 m
Answer: maximum height reached = 0.72 m

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