USE CORRECT UNITS A 14.0 kg block is dragged over a rough, horizontal surface by

Question

USE CORRECT UNITS

A 14.0 kg block is dragged over a rough, horizontal surface by a 77.0 N force acting at 20.0° above the horizontal. The block is displaced 3.00 m, and the coefficient of kinetic friction is 0.300.(a) Find the work done on the block by the 77.0 N force.

(b) Find the work done on the block by the normal force.

(c) Find the work done on the block by the gravitational force.

(d) What is increase in internal energy of the block-surface system due to friction?

(e) Find the total change in the block's kinetic energy.

Explanation / Answer

(a) by W = F in the direction of displacement x displacement
=>W = F x cosA* x s
=>W = 77 x cos20* x 3
=>W = 217.0689954 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 3 x [mg - Fsin20*]
=>delta E = 0.3 x 3 x [14 x 9.8 - 77 x 0.34]
=>delta E = 99.918 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 = 217.0689954 - 99.918 = 117.1509954 J

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