Question

A 575-g block is dropped onto a relaxed vertical spring that has a spring constant k =200.0 N/m. The block becomes attached to the spring and compresses the spring 58.1 cm before momentarily stopping. While the spring is being compressed, what work is done on the block by the gravitational force on it?What is the speed of the block just before it hits the spring? (Assume that friction is negligible.)

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

Answer 1

part 1:

work done on the block by grivational force=gravitational force*distance compressed

=0.575*9.8*0.581=3.274 J

part 2:

let the speed of the block just before hitting the spring be v m/s

then initial kinetic energy=0.5*0.575*v^2=0.2875*v^2

in the final scenario, when the block stops momentarily,

potential energy of the block=-mass*g*height (as the height is negative)

=-0.575*9.8*0.581=-3.274 J

potential energy of the spring=0.5*k*compression^2=0.5*200*0.581^2=33.7561 J

kinetic energy of the block=0 (as speed=0)

hence using conservation of energy principle:

initial total energy=final total energy

==>0.2875*v^2=-3.274+33.7561

==>0.2875*v^2=30.4821

==>v=10.2968 m/s

part 3:

speed at impact=2*10.2968=20.5936 m/s

kinetic energy of the block just before impact=0.5*0.575*20.5936^2=121.9277 J

then let compression be d.

then final potential energy of the block=-0.575*9.8*d

final potential energy of the spring=0.5*200*d^2

using conservation of energy principle as applied in part 2:

121.9277=-0.575*9.8*d+0.5*200*d^2

==>100*d^2-5.635*d-121.9277=0

solving for d,

we get d=1.1327 m

hence maximum compresion achieved is 1.1327 m