.. .. A 5.0 Kg block is released from point A in the figure below. It travels al

Question

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A 5.0 Kg block is released from point A in the figure below. It travels along the smooth surface A rightarrow E except for the part B rightarrow C where the surface is rough. Point A is 3 m above the lowest point C. The block continues in its motion up the surface until reaches point D (1.5 m above C) where it starts compressing the spring until it comes momentarily to rest at point E ( delta x above point D). The speed of the block at point C is 6 m/s. calculate the energy lost due to friction in the part B rightarrow C; if the spring constant kspring = 340 N/m, calculate the distance (delta x) by which the mass compresses the spring just before it comes to rest momentarily.

Explanation / Answer

Here at the Velocity at C = 6 m/sec


Here Kinetic Energy of the block at C = Potential Energy Gained from Point C to E + Elastic Potential Energy gained by the Spring

Therefore

0.5*5*6^2 = 5*9.8*(1.5+x) + 0.5*340*x^2

x = 0.1991 m = 19.91 cm

Energy Lost due to Friction = Potential Energy at A - Kinetic Energy at C

= 5*9.8*3 - 0.5*5*6^2

= 57 J

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