Question

A) A block of mass, m, is pushed up against a spring, compressing it a distance x, and is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case?

B) Initially a body moves in one direction and has kinetic energy K. Then it moves in the opposite direction with three times its initial speed. What is the kinetic energy now?

Answer 1

a)

KE=1/2mv^2

F= -kx

KE= -kx

-kx= 1/2mv^2

when u quadruple the m, the x also quadruples. when the speed is multiplied by 3, since it is v^2, the x is multiplied by 3^2 , which is 9

so distance the spring compressed is x×4×9= 36x

b) using the KE formular too,

when the v is tripled, since formula is v^2, the resulting KE will be multiplied by 3^2, so KE now= 9K