A block is pushed against the spring with spring constant 11 kN/m (located on th

Question

A block is pushed against the spring with spring constant 11 kN/m (located on the left-hand side of the trade) and compresses the spring a distance 5.3 cm from its equilibrium position (as shown in the figure below). The block starts at rest, is accelerated by the compressed spring, and slides across a frictionless track except for a small rough area on a horizontal section of the track (as shown in the figure below). It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s^2. What is the speed t' of the block when it leaves the track?

Explanation / Answer

here,

spring constnat , K = 11 KN/m = 11000 N/m

spring compression distance , x = 0.053 m

u = 0.4

r = 1 m

the work done by friction = the potential energy stored in the spring - kinetic energy gained

u * m * g * r = 0.5 * k * x^2 - 0.5 * m * v^2

0.4 * 0.581 * 9.81 * 1 = 0.5 * 11000 * 0.053^2 - 0.5 * 0.581 * v^2

v = 6.73 m/s

the speed v of the block when it leaves the track is 6.73 m/s

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