A block of mass 2 kg slides on a rough surface at t=0, if speed is 2 m/s. It stops after covering a distance of 20 cm because of friction. Find work done by the friction.

Open in App

Solution

initial Kinetic energy , $K_{1} = \frac{1}{2} m v_{1}^{2} = \frac{1}{2} \times 2 \times 2 = 2 J$

Final Kinetic energy , $K_{2} = 0$

Using work energy theorem we get
Work done by friction= Change in kinetic energy = $K_{2} - K_{1} = - 2 J$

Suggest Corrections

Similar questions

Q. A block of mass

2.0 kg

slides on a rough surface. At

t = 0

, its speed is

2.0 m/s

. It stops after covering a distance of

20 cm

because of the friction exerted by the surface on it. Find the work done by friction.

A block of mass 10 kg slides with velocity of 10 m/s on a rough horizontal surface. It comes to rest after covering a distance of 50 meters. Find the frictional force.

Q. A block of mass

60

kg just slides over a horizontal distance of

0.9

m. If the coefficient of friction between their surface is

0.15

then work done against friction will be

Q. A block of mass

50 k g

slides over a horizontal distance of

1 m

. If the coefficient of friction between their surfaces is

0.2

,then work done against friction is:

A block of mass 10 kg slides with velocity of 10 m/s on a rough horizontal surface. It comes to rest after covering a distance of 50 meters. Find the frictional force.

Join BYJU'S Learning Program

A block of mass 2 kg slides on a rough surface at t=0, if speed is 2 m/s. It stops after covering a distance of 20 cm because of friction. Find work done by the friction.

initial Kinetic energy , K1=12mv21=12×2×2=2 JFinal Kinetic energy , K2=0Using work energy theorem we get Work done by friction= Change in kinetic energy = K2−K1=−2 J

initial Kinetic energy , $K_{1} = \frac{1}{2} m v_{1}^{2} = \frac{1}{2} \times 2 \times 2 = 2 J$

Final Kinetic energy , $K_{2} = 0$

Using work energy theorem we get
Work done by friction= Change in kinetic energy = $K_{2} - K_{1} = - 2 J$