A worker pushed a 27kg block 9.2 m along a level floor

atconstant speed with a force directed 32o below thehorizontal. If

the coefficient of kinetic friction betweenblock and floor was

0.20, what were a) the work done by theworker's force and b) the

increase in thermal energy of theblock-floor system?

a) 557 J b) 557 J

Explanation:

The force acting on the body has vertical and horizontal component

the vertical component of the acting on the body = F sinθ

the horizontal component of the force acting on the block = F cosθ

since the block is moving with constant speed

then the frictional force = F cosθ in the opposite direction

μk = frictional force / force of normal

force of normal = mg + f sinθ

μk (mg + Fsinθ) = F cosθ

μkmg = F cosθ - μk Fsinθ = F (cos 32 - 0.2sin32) = 0.2 * 9.81 * 27

F (0.848 - 0.10598) = 52.974

F = 52.974 / (0.848 - 0.10598) = 71.392 N

A) work done = Fd cosθ = 71.392 * 9.2 * cos32 = 557 J

B) the increase in thermal energy of the block-floor system = 557J since the frictional force is equal to the force in the horizontal direction.