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

Answer:

k = 513.7 N/m

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

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

$W = (0+\dfrac{1}{2}kx^2) - (\dfrac{1}{2}mv^2 +0)$.........(2)

equating equation (1) and (2)

$-\mu mg x = \dfrac{1}{2}kx^2-\dfrac{1}{2}mv^2 +0)$

$k = \dfrac{-2\mu mg x + m v^2}{x^2}$

$k = \dfrac{-2\times 0.56\times 1.8 \times 9.8 \times 0.13+1.8\times 2.5^2}{0.13^2}$

k = 513.7 N/m

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