Question
A box slides along a rough floor with initial kinetic energy of 7 J. A person pushes on the box, in the same direction as its velocity doing 25 J of work. The velocity of the box is doubled. Find the magnitude of the work done by friction.
Answers
bobpursley
initial KE=7
final KE=7+25-frictionwork
velocity is doubled, so KE is quadrupled.
final KE=28J
28=7+25-friction
solve for friction
final KE=7+25-frictionwork
velocity is doubled, so KE is quadrupled.
final KE=28J
28=7+25-friction
solve for friction