Question

the work done in slowly pulling a body of mass 5 kg along an inclined plane of angle 60° with coefficient of friction 0.2 through 2 metre will be what

Answer 1

In case of the Friction, acceleration will be given by,

a  = gsin60 - μgcos60

a = 10√3/2 - 0.2 × 10 × 1/2

⇒ a = 5√3 - 1

Now,

Force = ma

= 5(5√3 - 1)

∴ Work done = 5(5√3 - 1) × 2

= 76.6 J.

Hope it helps.  

Answer 2

Answer:

In case of the Friction, acceleration will be given by,

a  = gsin60 - μgcos60

a = 10√3/2 - 0.2 × 10 × 1/2

⇒ a = 5√3 - 1

Now,

Force = ma

= 5(5√3 - 1)

∴ Work done = 5(5√3 - 1) × 2

= 76.6 J.