A 200- g block ispressed against a spring of force constant 1.40 kN/ m until the

Question

A 200- g block ispressed against a spring of force constant 1.40 kN/ m until theblock compresses the spring 10.0 cm. The spring rests at the bottomof a ramp inclined at 60.0° to the horizontal. Using energyconsiderations, determine how far up the incline the block movesbefore it stops ( a) if the ramp exerts no friction force on theblock and ( b) if the coefficient of kinetic friction is0.400. A 200- g block ispressed against a spring of force constant 1.40 kN/ m until theblock compresses the spring 10.0 cm. The spring rests at the bottomof a ramp inclined at 60.0° to the horizontal. Using energyconsiderations, determine how far up the incline the block movesbefore it stops ( a) if the ramp exerts no friction force on theblock and ( b) if the coefficient of kinetic friction is0.400.

Explanation / Answer

the potential energy stored in the spring before releaseis: PE = (1/2)kx2 =(1/2)*1.40kN*(0.10m)2 = 7.0J a) let the height it reach is H, and distance is S, we have H =S*sin(60) for energy conservation, mgH = PE = 7.0J =>mgS*sin(60) = 7.0J S=7.0J/(mgsin(60)) = 7.0J/(0.200kg*9.8m/s2*sin(60))= 4.12 m b) now the frction force is -mgcos(60), and work done by itis W=-mgcos(60)*S we have: mgH= PE - mgcos(60)*S =>mgSsin(60) + mgcos(60)S = PE = 7.0J S = 7.0J/[mg*(sin(60)+0.400*cos(60))] = 3.35m

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