You drop a brick of mass 2.2 kg from a height of 1.2 m above a spring of stiffne

Question

You drop a brick of mass 2.2 kg from a height of 1.2 m above a spring of stiffness 2100 N/m.

(a) How fast is the brick moving just before it hits the top of the spring?

(b) What is the maximum distance by which the spring is compressed?

(c) Up until now, you should have ignored any effects of friction. Now, though, assume that the brick bobs up and down on the spring for a while, slowly coming to rest because of dissipative forces inside the spring. How much is the spring compressed once this equilibrium is reached?

Explanation / Answer

a. For, speed of the brick(v)

v = sqrt ( 2 * g * h)

= sqrt( 2*9.81*1.2)

=4.85 m/s.

b. Conservation of energy

m g h = (- mg x) + 0.5 k x^2 +0(KE)

2.2 * 9.81 * 1.2 = - 2.2 * 9.81* x + .5(2100) x^2,

1050 x^2-21.58 x-25.89 = 0

x = -.147 or .167

Negative value is not possible,

x =.167m

c.

k*x' = m* g

x' = m*g/k

= 2.2*9.81/2100

=0.010 m

= 1.0 cm

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