An m = 3.38kg block situated on a rough incline is connected to a spring of negl

Question

An m = 3.38kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of k = 100N/m, as seen in the figure below. The pulley is frictionless. The block is released from rest when the spring is unstretched. The block moves 15.5cm down the incline before coming to rest. Assume ? = 38.4

Explanation / Answer

Let m = mass of block. Normal reaction from the incline = mg cos 38.4° = (0.78) mg So frictional force = (0.78) 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 (0.155 sin 38.4°) = (0.78) µ mg (0.155) + (0.5)k(0.155)^2, where k = spring constant => mg (sin 38.4°) = (0.78) µ mg + (0.5)k(0.155) => 6.08m = 7.644 m µ + 0.0775 k => µ = (6.08m - 0.07755 k) / (7.644 m). => µ=-.217 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 15.5 cm is not possible in the given condition.

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