Work Energy And Power MCQs for NEET — Physics Questions with Answers

The NCERT text points out that for conservative forces, 'the work done by this force in a closed path is zero.' Conversely, for non-conservative forces like friction, 'work done by friction over a closed path is not zero and no potential energy can be associated with friction.' This directly indicates that the work done by a non-conservative force over a closed path is generally non-zero.

The NCERT exercise asks about 'work done by the resistive force of air on a vibrating pendulum in bringing it to rest,' implying it does negative work, causing energy loss. The text explicitly states 'work done by friction or viscous force on a moving body is negative.' When a non-conservative force like air resistance does negative work, it causes a reduction in the system's total mechanical energy, leading to damping and eventually bringing the pendulum to rest.

The general form of the Work-Energy Theorem states $\Delta K = W_{net}$. The NCERT text elaborates: 'By the WE theorem ($F_c + F_{nc}) \Delta x = \Delta K$.' Here, ($F_c + F_{nc}$) represents the net force, so $\Delta K$ is equal to the total work done by all forces (conservative and non-conservative).

The NCERT exercises ask to state the sign of work done by various forces. One option is 'work done by friction on a body sliding down an inclined plane'. In such a scenario, friction acts opposite to the direction of motion, thus doing negative work. The text also generally states 'The work done by the friction or viscous force on a moving body is negative.' Gravity, spring force, and electrostatic force are conservative forces and can do positive or negative work, but friction is explicitly non-conservative and its work is often negative during motion.

The NCERT text states: 'The total mechanical energy of a system is conserved if the forces, doing work on it, are conservative.' Conversely, if mechanical energy is not conserved, it implies that non-conservative forces are doing work on the system. The equation $\Delta E = W_{nc}$ directly shows that a change in total mechanical energy ($\Delta E$) is due to the work done by non-conservative forces ($W_{nc}$). While conservative forces may still be present, it's the non-conservative ones that cause non-conservation of mechanical energy.

For uniform velocity, the net force on the body is zero. This means 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, doing positive work. The frictional force acts opposite to the direction of motion, doing negative work. Since the body moves with uniform velocity, the work done by the applied force must be equal in magnitude to the negative work done by friction, so that net work done is zero, and thus change in kinetic energy is zero (as velocity is constant). The NCERT exercise mentions 'work done by an applied force on a body moving on a rough horizontal plane with uniform velocity' and 'work done by friction on a body sliding down an inclined plane' (which implies negative work for friction). Hence, the applied force does positive work, and friction does negative work of equal magnitude.

According to the NCERT text, 'Work done is a scalar quantity. It can be positive or negative unlike mass and kinetic energy which are positive scalar quantities.' Potential energy is undetermined up to a constant and can be positive, negative, or zero depending on the choice of reference point. Displacement is a vector quantity and its component can be positive or negative.

As per the NCERT text, 'if the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 5.3(b). Then the work done is $W = \int_{x_i}^{x_f} F(x) dx$'. This integral represents the area under the force-displacement curve.

According to Example 5.5, the work done by the woman for the first 10 m (rectangle ABCD) is $100 \text{ N} \times 10 \text{ m} = 1000 \text{ J}$. For the next 10 m (trapezium CEID), the work done is $(1/2) \times (100 + 50) \text{ N} \times 10 \text{ m} = 750 \text{ J}$. The total work done by the woman is $1000 \text{ J} + 750 \text{ J} = 1750 \text{ J}$.

As per Example 5.5, the frictional force $f$ is constant at 50 N and opposes motion, hence it is negative. The work done by friction is $W_f = (-50 \text{ N}) \times 20 \text{ m} = -1000 \text{ J}$. The area on the negative side of the force axis has a negative sign.