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. 


1
Expert's answer
2021-10-26T09:43:58-0400

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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