A19.6 kg block is dragged over a rough, horizontal surface by a constant force of 148 n acting at an angle of 30.7 ◦ above the horizontal. the block is displaced 64.7 m, and the coefficient of kinetic friction is 0.185. 19.6 kg µ = 0.185 148 n 30.7 ◦ find the work done by the 148 n force. the acceleration of gravity is 9.8 m/s 2 . answer in units of j. 003 (part 2 of 2) 10.0 points find the magnitude of the work done by the force of friction. answer in units of j.

(a) The total work done by the 148 N force is 6838.92 J

(b) The magnitude of the work done by the force of friction is -1394.68 J

Explanation:

Here, we have, resolving the 148 N into vertical, Fv and horizontal, Fh components

Fv = 148 × sin 30.7 = 75.56 N

Fh = 148 × cos 30.7  = 127.26 N

Force of friction = Normal Force, F(N) × Coefficient of Friction, μ

Where:

Coefficient of Friction, μ = 0.185

Normal Force = 19.6 × 9.8 - 148 × sin 30.7 =  116.52 N

Force of friction = 116.52 N × 0.185 = 21.56 N Horizontal negative

Total Horizontal force required to pull = 127.26 N - 21.56 N = 105.7 N

The work done in pulling along the horizontal  is given by

Work done = Magnitude of Horizontal Force × Distance of force

Work done to pull = 105.7 N  × 64.7 m = 6838.92 J

While the work done less frictional work = 8233.6 J

(b) The magnitude of work done against friction =

Force of friction × Distance moved

= 21.56 N × 64.7 m = 1394.68 J

The magnitude of work done by frictional force = -1394.68 J.

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures