Question

9. A 1.80 kg block slides on a rough horizontal surface. The block hits a spring with a speed of 2.5m/s and compresses it a distance of 13cm before coming to rest. If the coefficient of kinetic friction between the block and surface is 0.560, what is the force constant of the spring? (Use non-conservative energy principle to solve this problem)

Answer 1

k = 513.7 N/m

Step-by-step explanation:

given,

mass of block = 1.80 Kg

Speed at which block hit spring = 2.5 m/s

compression length(x) = 0.13 m

coefficient of kinetic friction = 0.56

force constant of the spring =?

non conservative work done by the frictional force

W = -f d

W = - μ mg x...............(1)

non conservation work done

`W = E_f - E_i`

Work done is change in energy

`W = (K_f+U_f) - (K_i + U_i)`

`W = (0+(1)/(2)kx^2) - ((1)/(2)mv^2 +0)`.........(2)

equating equation (1) and (2)

`-\mu mg x = (1)/(2)kx^2-(1)/(2)mv^2 +0)`

`k = (-2\mu mg x + m v^2)/(x^2)`

`k = (-2* 0.56* 1.8 * 9.8 * 0.13+1.8* 2.5^2)/(0.13^2)`

k = 513.7 N/m

force constant for spring is equal to k = 513.7 N/m