across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 165 N directed 26 degrees below the horizontal. While you slide the chair a distance of 6.00 m, the chair's speed changes from 1.30 m/s to 2.50 m/s. Find the work done by friction on the chair.

-847.2J

Explanation:

First find the acceleration from v^2 = u^2 + 2as

v = 2.5 m/s

u = 1.3 m/s

a?

s=6.00

a = v^2-u^2/2s

a = (2.5) ^2 - (1.3) ^2/2 * 6

a = 0.38ms^-2

From Newtons second law:

(Force applied cos Θ) - (Frictional force) = ma

Frictional force = ma - (Force applied cos Θ)

Frictional force = (18.8*0.38) - (165 cos 26°)

Frictional force = 7.144 - 148.3 = - 141.2N

Therefore,

Work done by friction = Frictional force * distance covered

= - 141.2N * 6 = - 847.2J

W = - 847J

Explanation:

Given m = 18.8kg, F = 165N, θ = - 26° (below the horizontal, s = 6.0m, u = 1.30m/s and v = 2.50m/s

In this problem, two forces act on the chair; the forward force F and the frictional force f. We would apply newton's second law to find the frictional force f after which we can calculate the workdone by the frictional force f*s.

But for us to apply newton's second law, we need to know the acceleration of the chair cause by the net force.

From constant acceleration motion equations

v² = u² + 2as

2.5² = 1.30² + 2a*6

6.25 = 1.69 + 12a

12a = 6.25 - 1.69

12a = 4.56a

a = 4.56/12

a = 0.38m/s

By newton's second law the net sum of forces equals m*a

The force F has horizontal and vertical and components. It is the horizontal component of this force that pushes the chair against friction.

Fx and f are oppositely directed.

So

Fx - f = ma

165cos (-26) - f = 18.8*0.38

148.3 - f = 7.14

f = 148.3 - 7.14

f = 141.2N

Workdone = - fs = - 141.2*6.00 = - 847J

W = - 847J

Work is negative because it is done by a force acting on the chair in a direction opposite (antiparallel) to that of the intended motion.