Question #88716
A 3 kg block is moved up a 37 degrees incline under the action of a constant horizontal force of 40 N. The coefficient of kinetic friction is 0.1, and the block is displaced 2 meters up the inclined. Calculate (a) the work done by the 40 N force,(b) the work done by the gravity, (c) the work done by the friction, and (d) the changed in kinetic energy of the block
1
Expert's answer
2019-07-04T11:10:59-0400

(a) The work done by the 40 N force:


W1=Fd=(40)(2)=80 JW_1=Fd=(40)(2)=80 \ J

(b) The work done by the gravity:


W2=mgh=mgdsin37=(3)(9.8)(2)sin37=35.4 JW_2=mgh=mgd \sin{37}=(3)(9.8)(2) \sin{37}=35.4\ J

(c) The work done by the friction:


W3=Ffd=μmgdcos37=0.1(3)(9.8)(2)cos37=4.7 JW_3=F_fd=\mu mgd \cos{37}=0.1(3)(9.8)(2) \cos{37}=4.7\ J

(d) ) The change in kinetic energy of the block:


ΔK=W1W2W3=8035.44.7=40 J\Delta K=W_1-W_2-W_3=80-35.4-4.7=40\ J



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