A 500-g block is dropped onto a vertical spring with spring constant k =140.0 N/

Question

A 500-g block is dropped onto a vertical spring with spring constant k =140.0 N/m. The block becomes attached to the spring, and the spring compresses 74.1 cm before momentarily stopping.

A) While the spring is being compressed, what work is done on the block by its weight?

B) While the spring is being compressed, what work is done on the block by the spring force?

C) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

D) If the speed at impact is doubled, what is the maximum compression of the spring?

Explanation / Answer

W = m*g = 0.5*9.8 = 4.9 N

K = 140 N/m

part A)

work = W*dy = 4.9*0.741 = 3.6309 J

part B)

Work done by spring = 0.5*K*dx^2 = 0.5*140*0.741*0.741 = 38.43567 J

part C)

just before impact KE of the block = 0.5*m*v^2

PE of the block = m*g*dy

total energy of the block = KE + PE = 0.5*m*v^2 + m*g*dy

after the block momentarily stops energy is stored in spring PE stored in spring Ue = 0.5*K*dx^2

0.5*m*v^2 + m*g*dy = 0.5*K*dx^2

0.5*0.5*v^2 + (0.5*9.8*0.741) = 0.5*140*0.741*0.741

v = 11.8 m/s <----answer

part D)

v' = 2v

initial TE = 0.5*m*(2v)^2 + m*g*dy'

final PE = 0.5*K*dy'^2

0.5*m*(2v)^2 + m*g*dy' = 0.5*K*dy'^2

(0.5*0.5*23.6*23.6) + (0.5*9.8*dy') = (0.5*140*dy'^2)

139.24 + 4.9dy' = 70*dy'^2

dy' = 1.445 m = 144.5 cm

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