A 300-g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 300-g block is dropped onto a relaxed vertical spring that has a spring constant k =140.0 N/m. The block becomes attached to the spring and compresses the spring 46.6 cm before momentarily stopping. While the spring is being compressed, the work done on the block by the gravitational force is 1.37 J. While the spring is being compressed, the work is done on the block by the spring force is -1.52×101 J. The speed of the block just before it hits the spring is 9.60 m/s

QUESTION: If the speed at impact is doubled, what is the maximum compression of the spring?

ANSWER: 9.10×10-1 m

HOW???? Please type, I have a hard time reading handwritting.

Explanation / Answer

mass of block(m)=0.3kg

spring constant(k)=140N/m

velocoity of ball=2times of initial velocity

=(2*9.6)m/s

let the maximum compression of spring be X

applying the law of conservation of energy we get

change in kinetic energy =work done spring force+work done by gravity

(1/2)Kx^2-Mgx=(1/2)m(2u)^2

(140/2)x^2-(0.3*10*x)=2*0.3*9.6^2

70X^2-3X-55.296=0

solving this equation we get

X=0.910506m

maximum compression in spring is 9.10*10^-1m

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