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

Let's analyze the sign of work done for each scenario:

1. Work done by a man lifting a bucket: The man applies an upward force to lift the bucket, and the bucket moves upward. Since the applied force and the displacement are in the same direction, the work done by the man is positive.
2. Work done by gravitational force when lifting a bucket: Gravity acts downwards, while the bucket is lifted upwards. The force of gravity and the displacement are in opposite directions. Therefore, the work done by the gravitational force is negative.
3. Work done by friction on a body sliding down an inclined plane: Friction always opposes the motion. As the body slides down, the frictional force acts upwards along the plane. Since the frictional force is opposite to the direction of displacement, 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 surface, the applied force must be equal in magnitude and opposite in direction to the opposing forces (like friction). If the applied force is in the direction of motion, and the motion is occurring, the work done by this 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, causing the pendulum to eventually come to rest.

Given:

• Mass of the body, $m\; =\; 2\; \backslash text\{\; kg\}$
• Applied horizontal force, $F\_\{app\}\; =\; 7\; \backslash text\{\; N\}$
• Coefficient of kinetic friction, $\backslash mu\_k\; =\; 0.1$
• Initial velocity, $u\; =\; 0\; \backslash text\{\; m/s\}$ (since it starts from rest)
• Time, $t\; =\; 10\; \backslash text\{\; s\}$
• Acceleration due to gravity, $g\; \backslash approx\; 9.8\; \backslash text\{\; m/s\}^2$

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

1. Normal force (N): Since the motion is horizontal, the normal force exerted by the table on the body is equal to the gravitational force acting on it.

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

2. Frictional force ($f\_k$): The kinetic friction opposes the motion.

$$f\_k\; =\; \backslash mu\_k\; N\; =\; (0.1)(19.6\; \backslash text\{\; N\})\; =\; 1.96\; \backslash text\{\; N\}$$

3. Net force ($F\_\{net\}$): The net force acting horizontally is the applied force minus the frictional force.

$$F\_\{net\}\; =\; F\_\{app\}\; -\; f\_k\; =\; 7\; \backslash text\{\; N\}\; -\; 1.96\; \backslash text\{\; N\}\; =\; 5.04\; \backslash text\{\; N\}$$

Now, let's calculate the motion of the body:

4. Acceleration (a): Using Newton's second law, $F\_\{net\}\; =\; ma$.

$$a\; =\; \backslash frac\{F\_\{net\}\}\{m\}\; =\; \backslash frac\{5.04\; \backslash text\{\; N\}\}\{2\; \backslash text\{\; kg\}\}\; =\; 2.52\; \backslash text\{\; m/s\}^2$$

5. Velocity after 10 s (v): Using the equation of motion, $v\; =\; u\; +\; at$.

$$v\; =\; 0\; +\; (2.52\; \backslash text\{\; m/s\}^2)(10\; \backslash text\{\; s\})\; =\; 25.2\; \backslash text\{\; m/s\}$$

Now we can compute the required quantities:

a) Work done by the applied force in 10 s ($W\_\{app\}$):

The applied force is $F\_\{app\}\; =\; 7\; \backslash text\{\; N\}$. The distance moved in 10 s ($d$) can be found using $d\; =\; ut\; +\; \backslash frac\{1\}\{2\}at^2$.

$$d\; =\; (0)(10)\; +\; \backslash frac\{1\}\{2\}(2.52\; \backslash text\{\; m/s\}^2)(10\; \backslash text\{\; s\})^2\; =\; \backslash frac\{1\}\{2\}(2.52)(100)\; \backslash text\{\; m\}\; =\; 126\; \backslash text\{\; m\}$$

Work done by the applied force is $W\_\{app\}\; =\; F\_\{app\}\; \backslash times\; d$ (since force and displacement are in the same direction).

$$W\_\{app\}\; =\; (7\; \backslash text\{\; N\})(126\; \backslash text\{\; m\})\; =\; 882\; \backslash text\{\; J\}$$

b) Work done by friction in 10 s ($W\_f$):

The frictional force is $f\_k\; =\; 1.96\; \backslash text\{\; N\}$ and it acts opposite to the direction of motion.

$$W\_f\; =\; f\_k\; \backslash times\; d\; \backslash times\; \backslash cos(180^\backslash circ)\; =\; (1.96\; \backslash text\{\; N\})(126\; \backslash text\{\; m\})(-1)\; =\; -246.96\; \backslash text\{\; J\}$$

c) Work done by the net force on the body in 10 s ($W\_\{net\}$):

The net force is $F\_\{net\}\; =\; 5.04\; \backslash text\{\; N\}$.

$$W\_\{net\}\; =\; F\_\{net\}\; \backslash times\; d\; =\; (5.04\; \backslash text\{\; N\})(126\; \backslash text\{\; m\})\; =\; 635.04\; \backslash text\{\; J\}$$

Alternatively, $W\_\{net\}\; =\; W\_\{app\}\; +\; W\_f\; =\; 882\; \backslash text\{\; J\}\; +\; (-246.96\; \backslash text\{\; J\})\; =\; 635.04\; \backslash text\{\; J\}$.

d) Change in kinetic energy of the body in 10 s ($\backslash Delta\; KE$):

Initial kinetic energy, $KE\_i\; =\; \backslash frac\{1\}\{2\}mu^2\; =\; \backslash frac\{1\}\{2\}(2\; \backslash text\{\; kg\})(0\; \backslash text\{\; m/s\})^2\; =\; 0\; \backslash text\{\; J\}$.

Final kinetic energy, $KE\_f\; =\; \backslash frac\{1\}\{2\}mv^2\; =\; \backslash frac\{1\}\{2\}(2\; \backslash text\{\; kg\})(25.2\; \backslash text\{\; m/s\})^2\; =\; (1)(635.04)\; \backslash text\{\; J\}\; =\; 635.04\; \backslash text\{\; J\}$.

Change in kinetic energy, $\backslash Delta\; KE\; =\; KE\_f\; -\; KE\_i\; =\; 635.04\; \backslash text\{\; J\}\; -\; 0\; \backslash text\{\; J\}\; =\; 635.04\; \backslash text\{\; J\}$.

Interpretation of results:

• The work done by the applied force is positive, indicating that this force has added energy to the system.
• The work done by friction is negative, indicating that friction has removed energy from the system.
• The work done by the net force is positive and is equal to the change in kinetic energy of the body. This confirms the work-energy theorem.
• The change in kinetic energy is positive, meaning the body has sped up, which is consistent with a net positive work done on it.