A 2.9 kg block is dropped from a height of 4.5 m onto a spring of spring constant 3955 N/m.?

Question

Find the speed of the block when the compression of the spring is 15.0 cm.

Answer 1

When the 2.9 kg block has dropped a total distance of 4.65 m, it has released
mgh = 2.9*9.8*4.65 = 132.153J of potential energy. The compressed spring contains
kx²/2 = 3655*.15²/2 = 41.119J of potential energy which means that
132.153 - 41.119 = 91.034J of energy must be the kinetic energy remaining in the block. So
mv²/2 = Ek => v = √(2*Ek/m)
√(2*91.034/2.9) = 9.541 m/s.

HTH ☺


Doug

Answer 2

initial energy of block when no compression = m*g*H

(potenetial energy line is set zero at at 15cm compression of srping )
= 2.9*9.81*(4.5 + 0.15)
= 132.28785 joules

work done on block by spring = - 0.5 k x^2
= - 0.5*3955*0.15^2
= -44.4937 joules

so net energy of block = 132.28785-44.4937
= 87.79415 joules

so kinetic energy = 0.5* m * v^2 = 87.79415 ( as poyential energy here is set zero)

0.5*2.9*v^2 = 87.79415
v = sqrt(60.769)
= 7.7812 m/s