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a boy of weight 25kg slide sdown a rope hanging from the branch of the tree.if the force of friction against him is 50N,the boys acceleration is?
Simply balance the forces along with friction force.Therfore,F – f = ma250 – 50 = 25aa = 200/25 = 40/5 = 8 m/s^2
by applying dyanamic equation F-fk=mamg-50=25a250-50=25a200/25=aso acceleration=8metre/second square.40/5=aa=8m/sec2
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