A 5.0-kg block is attached to an ideal massless spring whose spring constant is

Question

A 5.0-kg block is attached to an ideal massless spring whose spring constant is 125 N/m. The block is pulled from its equilibrium position at x= 0.00 m to a position at x= +0.687 m and is released from rest. The block then executes lightly damped oscillation along the x-axis, and the damping force is proportional to the velocity. When the block first returns to x= 0.00 m, its x component of velocity is -2.0 m/s and its x component of acceleration is +5.6 m/s2.

(a) What is the magnitude of the acceleration of the block upon release at x= +0.687 m?

(b) Find the damping constant b.

Explanation / Answer

a)

m = mass = 5 kg

k = spring constant = 125 N/m

X = stretch in the spring = 0.687 m

the force on the block when the spring is stretched

F = kX

ma = kx         (since F = ma)

a = (125) (0.687) / (5)

a = 17.175 m/s2

b)

the equation of motion of block under damping is given as ::

ma = -kx - bV

at x =0 , V = -2 m/s , a = +5.6 m/s2

(5) (5.6) = - (125) (0) - b (-2)

b = 14

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