please explain thanks A block is pushed against the spring with spring constant

Question

please explain thanks

A block is pushed against the spring with spring constant k (located on the left-hand side of the track) and compresses the spring a distance 4.8 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 9.81 m/s2. What is the spring constant k? Answer in units of N/m

Explanation / Answer

range of the block = 4.17 = vt

height from which spring is thrown = 2.2 = .5*g*t^2 where g is acc. due to gravity and t is time taken to reach ground from the time it is projected from the edge of the wall

then from above equations, v^2 / (.5*g) = 4.17^2 /2.2, then v = 6.23 m/s

energy given by the spring = .5k*x^2 = .5*k*(.048)^2 = k*1.152*10^-3 J

energy lost due to friction = u*m*g*s = .4*.535*9.81*1.1 = 2.31 J

energy of the block at the edge of the wall when it is projected from it = .5*.535*6.23^2 = 10.38J

then k*1.152*10^-3 = 2.31 + 10.38, then k = 11.014 kN/m

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