Question

A particle moves on a rough horizontal ground with some initial velocity V0. If 34th of its kinetic energy is lost in friction in time t0 the coefficient of friction is

Answer 1

Answer:

A

2gt

0

v

0

Initial Kinetic Energy =

2

1

mv

0

2

Let final velocity is v, ⇒ Final Kinetic Energy=

2

1

mv

2

=

2

1

mv

0

2

−

4

3

×

2

1

mv

0

2

⇒v=

2

v

0

Using v=u+at, Acceleration a=−

2t

0

v

0

Retardation, a=

2t

0

v

0

Let friction coefficient is μ ⇒a=μg

⇒μ=

2gt

0

v

0