Question

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.

Answer 1

Answer:

-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

Answer 2

Answer:

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.