A 13.9 kg block is dragged over a rough, horizontal surface by a 70.7 N force acting at 19.8 degrees above the horizontal. The block is displaced 4.55 m, and the coefficient of kinetic friction is 0.298.

Question

How much energy is lost due to friction?

Answer 1

break the 70.7N force into vertical and horizontal components.
The vertical component reduces weight, so figure the normal force (and friction) from that.

frictionforce=(13.9g-70.7Sin19.8)mu
work=frictionforce*distance

Now notice that this does not equal the horizontal component of force*distance. So, the block has to accelerate.