Question

A body has a mass of 3 kg which when applied with a force of 8 N moves from rest with a coefficient of kinetic friction = 0.2. Find the following:

(a) When a force is applied for 10 s, what is the work done?

(b) The work done by the friction in 10 s.

(c) When a net force acts on the body for 10 s, what is the work done?

(d) In the time interval of 10 s, the change in the kinetic energy.

Answer 1

Explanation:

Mass m = 3 kg

Force F = 8 N

Kinetic friction coefficient (μ) = 0.2

Initial velocity , u = 0

t = 10 s

According to the Newton’s law of motion:

{a}' =   $\frac{F}{m}$  

=$\frac{8}{3}$

​= 2.6 m/s2

Friction force =μmg

= 0.2 x 3 x -9.8 = -5.88 N

Acceleration due to friction:

{a}” =  $\frac{-5.88}{3}$

= -1.96 m/s2

The total acceleration of the body = {a}'+ {a}”

= 2.6 + (-1.96) = 0.64 m/s2

According to the equation of the motion

s = ut + $\frac{1}{2}$ at^2

= 0 + \frac{1}{2}  

​  

= 1/2   x 0.64 x (10)2

= 32 m

(a). Wa = F x s = 8 x 32 = 256 J

(b). Wt = F x s  = -5.88 x 32 = -188 J

Net force = 8 + (-5.88) = 2.12 N

(c). Wnet = 2.12 x 32 = 68 J

(d) Final velocity v = u + at

= 0 + 0.64 x 10 = 6.4 m/s

Change in kinetic energy = $\frac{1}{2}$ mv2 -$\frac{1}{2}$ mu2

=$\frac{1}{2}$ x 2 (v2 – u2)

= 6.42 – 02

= 41 J

Answer 2

Answer:

Explanation:Explanation:Explanation:Explanation: