A 2.00 kg block situated on a rough incline is connected to a spring of negligib

Question

A 2.00 kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m (Fig. P5.76). The block is released from rest when the spring is unstretched, and the pulley is frictionless. The block moves20.2cm down the incline before coming to rest. Find the coefficient of kinetic friction between the block and incline.

Explanation / Answer

Let m = mass of block. Normal reaction from the incline = mg cos 37° = (0.8) mg So frictional force = (0.8) µ mg Best approach to find the value of µ will be the energy balance. Loss of P.E. = Work done against friction + work done in stretching the spring. => mg (0.25 sin 37°) = (0.8) µ mg (0.25) + (0.5)k(0.25)^2, where k = spring constant => mg (sin 37°) = (0.8) µ mg + (0.5)k(0.25) => 5.9m = 7.84 m µ + 0.125 k => µ = (5.9m - 0.125 k) / (7.84 m). After submitting as above, I realized that values of m and k are given, but µ works out negative on substitution. I rechecked my steps and could not find any error. This means that streching by 25 cm is not possible in the given condition.

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