A body of mass 20 kg is resting on a rough horizontal plane of the co-effcient of friction 0.5. If 40 N force is applied on a body then find the velocity of the body after 10 sec. (Take g = 10 m/sec²)

CONCEPT:

Friction:

• The resistance offered by the surfaces that are in contact with each other when they move over each other is called friction.
• Factors affecting friction:
  1. Types of surfaces in contact.
  2. The normal force between the two surfaces.
• Friction force does not depend on the velocity of the object.

Limiting friction:

• The maximum friction that can be generated between two static surfaces in contact with each other.
• Once a force applied to the two surfaces exceeds the limiting friction, the motion will occur.
• We know that the limiting friction force between any two surfaces is given as,

⇒ F = μN

Where F = friction force, μ = coefficient of friction and N = normal reaction

Newton's second law of motion:

• According to Newton's second law of motion, the rate of change of momentum of a body is directly proportional to the applied unbalanced force.
• The magnitude of the force is given as,

⇒ F = ma

Where m = mass and a = acceleration

CALCULATION:

Given:

 m = 20 kg, μ = 0.5, P = 40 N, t = 10 sec and g = 10 m/sec2

​Since the body is resting on a rough horizontal plane so the normal reaction is given,

⇒ N = mg

⇒ N = 20 × 10

⇒ N = 200 N

The limiting friction is given as,

⇒ F = μN

⇒ F = 0.5 × 200

⇒ F = 100 N

• Since the applied force is less than the limiting friction so the body will remain at rest. Hence, option 3 is correct.