CBSE Class 11 Physics Chapter 6: Work, Energy and Power NCERT Solutions

The sign of work done depends on the angle between the force and the displacement.

1. Work done by the man lifting a bucket: The man applies an upward force on the bucket, and the bucket moves upward. Since the force and displacement are in the same direction, the work done by the man is positive.
2. Work done by gravitational force: The gravitational force acts downward on the bucket, while the displacement is upward. Since the force and displacement are in opposite directions, the work done by gravity is negative.
3. Work done by friction on a body sliding down an inclined plane: Friction always opposes motion. As the body slides down, the frictional force acts upward along the plane. The displacement is downward along the plane. Since the force and displacement are in opposite directions, the work done by friction is negative.
4. Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity: To maintain uniform velocity on a rough horizontal plane, the applied force must be equal in magnitude and opposite in direction to the frictional force. The applied force acts in the direction of motion, and the displacement is also in the same direction. Therefore, the work done by the applied force is positive.
5. Work done by the resistive force of air on a vibrating pendulum: The resistive force of air always opposes the motion of the pendulum. As the pendulum swings, this force acts in the direction opposite to its velocity. Therefore, the work done by the air resistance is negative.

Given:\nMass of the body, m = 2 \text{ kg}\nApplied horizontal force, F_{app} = 7 \text{ N}\nCoefficient of kinetic friction, \mu_k = 0.1\nInitial velocity, u = 0 \text{ m/s} (since it starts from rest)\nTime, t = 10 \text{ s}

First, let's calculate the forces acting on the body.

The normal force (N) exerted by the table on the body is equal to the weight of the body:

N = mg = (2 \text{ kg})(9.8 \text{ m/s}^2) = 19.6 \text{ N}

The force of kinetic friction (f_k) opposes the motion:

f_k = \mu_k N = (0.1)(19.6 \text{ N}) = 1.96 \text{ N}

Now, let's find the net force acting on the body in the horizontal direction:

F_{net} = F_{app} - f_k = 7 \text{ N} - 1.96 \text{ N} = 5.04 \text{ N}

Using Newton's second law (F_{net} = ma), we can find the acceleration (a) of the body:

a = \frac{F_{net}}{m} = \frac{5.04 \text{ N}}{2 \text{ kg}} = 2.52 \text{ m/s}^2

Now we can calculate the required quantities:

1. Work done by the applied force in 10 s:\nThe applied force is F_{app} = 7 \text{ N}. We need to find the displacement (d) in 10 s. Using the equation of motion d = ut + \frac{1}{2}at^2:

d = (0)(10) + \frac{1}{2}(2.52 \text{ m/s}^2)(10 \text{ s})^2 = \frac{1}{2}(2.52)(100) \text{ m} = 126 \text{ m}\nWork done by applied force, W_{app} = F_{app} \times d = (7 \text{ N})(126 \text{ m}) = 882 \text{ J}

2. Work done by friction in 10 s:\nThe frictional force is f_k = 1.96 \text{ N}, acting opposite to the displacement.\nWork done by friction, W_k = -f_k \times d = -(1.96 \text{ N})(126 \text{ m}) = -246.96 \text{ J}

3. Work done by the net force on the body in 10 s:\nWork done by net force, W_{net} = F_{net} \times d = (5.04 \text{ N})(126 \text{ m}) = 635.04 \text{ J}\nAlternatively, W_{net} = W_{app} + W_k = 882 \text{ J} + (-246.96 \text{ J}) = 635.04 \text{ J}

4. Change in kinetic energy of the body in 10 s:\nFirst, find the final velocity (v) after 10 s using v = u + at:

v = 0 + (2.52 \text{ m/s}^2)(10 \text{ s}) = 25.2 \text{ m/s}\nInitial kinetic energy, KE_i = \frac{1}{2}mu^2 = \frac{1}{2}(2 \text{ kg})(0 \text{ m/s})^2 = 0 \text{ J}\nFinal kinetic energy, KE_f = \frac{1}{2}mv^2 = \frac{1}{2}(2 \text{ kg})(25.2 \text{ m/s})^2 = (25.2)^2 \text{ J} = 635.04 \text{ J}\nChange in kinetic energy, \Delta KE = KE_f - KE_i = 635.04 \text{ J} - 0 \text{ J} = 635.04 \text{ J}

Interpretation of results:

- The work done by the applied force is positive, meaning energy is transferred to the system by this force.

- The work done by friction is negative, indicating that friction removes energy from the system.

- The work done by the net force is positive and is equal to the change in the kinetic energy of the body. This confirms the work-energy theorem.

- The change in kinetic energy is positive, as the body's speed increases from rest.