Answer to Question #253945 in Mechanics | Relativity for Kristine

Question

A 15.0-kg block is dragged over a rough, horizontal surface by a 70.0-N force acting at 20.0 o above the horizontal. The block is displaced at 5.00 m, and the coefficient of kinetic friction is 0.300. Find the work done by  a) the 70-𝑁 force, b) the normal force, and c) the force of gravity, d) What is the energy loss due to friction? e) Find the total change in the block’s kinetic energy. 

Answer 1

a) The work is

$$W=Fl\cos\theta=70·5·\cos20°=329\text{ J}.$$

b) The work is

$$W=Nl\cos\theta=NL\cos90°=0.$$

c) The work is zero like in (b).

d) The energy loss is

$$Q=\mu Nl=\mu(mg-F\sin20°)·l=\\=0.3·(15·9.8-70\sin20°)·5=185\text{ J}.$$

e) The change is

$$\Delta KE=W-Q=144\text{ J}.$$