A 15.0-kg block is dragged over a rough, horizontal surface by a 70.0 N force ac

Question

A 15.0-kg block is dragged over a rough, horizontal surface by a 70.0 N force acting 20.0 degrees above the horizontal. What is the magnitude of the change in the block's kinetic energy if the block is displaced 5.00 m and the coefficient of friction is .300?

Explanation / Answer

(a) by W = F in the direction of displacement x displacement =>W = F x cosA* x s =>W = 90 x cos20* x 7 =>W = 592 J (b) W by Normal force = 0, as no displacement in the direction of N take place, i.e. =>W = N x s =>W = N x 0 = 0 (c) Again 0, with same reason of (b) (d) By work energy theorem:- =>delta E = W =>delta E = W(friction) =>delta E = Ff x s =>delta E =coefficient of kinetic friction x N x s =>delta E = 0.3 x 7 x [mg - Fsin20*] =>delta E = 0.3 x 7 x [15 x 9.8 - 90 x 0.34] =>delta E = 244.06 J (e)delta KE = F(net) x s =>delta KE = F x cos20* x s - coefficient of kinetic friction x N x s =>delta KE = 592 - 244.06 = 347.94 J

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