Question #253945

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. 


Expert's answer

a) The work is


W=Flcosθ=705cos20°=329 J.W=Fl\cos\theta=70·5·\cos20°=329\text{ J}.

b) The work is


W=Nlcosθ=NLcos90°=0.W=Nl\cos\theta=NL\cos90°=0.

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

d) The energy loss is


Q=μNl=μ(mgFsin20°)l==0.3(159.870sin20°)5=185 J.Q=\mu Nl=\mu(mg-F\sin20°)·l=\\=0.3·(15·9.8-70\sin20°)·5=185\text{ J}.

e) The change is


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


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