Question

A 2kg block is released from the top of a 37º incline at a distance of 0.4 m from a light, unstretched spring. At the end of its slide, the block compresses the spring by 0.1m. If the coefficient of kinetic friction between the incline is 0.2 determine the spring constant.

The answer is 880N/m how does one find this answer thanks for your help.

Answer 1

here,

mass of block , m = 2 kg

theta = 37 degree

s = 0.4 m

compression in the spring , x = 0.1 m

coeffiicent of friction , uk = 0.2

let the spring constant be K

using work energy theorm

work done by friction = initial potential energy - final elastic potential energy

uk * m * g * (s + x) * cos(theta) = m * g * sin(theta) * ( s + x) - 0.5 * K * x^2

0.2 * 2 * 9.81 * ( 0.4 + 0.1) * cos(37) = 2 * 9.81 * sin(37) * ( 0.4 + 0.1) - 0.5 * K * 0.1^2

solving for K

K = 880 N/m

the spring constant is 880 N/m

answered by: anonymous