Work, Energy and Power — JEE Main practice

1. The force acting on a particle equals the negative slope of its potential energy versus position graph.

$$F = -\frac{dV}{dx}$$

2. So the magnitude of the force in each region equals the absolute value of the slope between that region's endpoints.

$$|F| = \left|\frac{V_2-V_1}{x_2-x_1}\right|$$

3. For region $AB$, from $(0,0)$ to $(1,1)$:

$$|F_{AB}| = \left|\frac{1-0}{1-0}\right| = 1$$

4. For region $BC$, from $(1,1)$ to $(2,4)$:

$$|F_{BC}| = \left|\frac{4-1}{2-1}\right| = 3$$

5. For region $CD$, from $(2,4)$ to $(4,4)$:

$$|F_{CD}| = \left|\frac{4-4}{4-2}\right| = 0$$

6. For region $DE$, from $(4,4)$ to $(6,3)$:

$$|F_{DE}| = \left|\frac{3-4}{6-4}\right| = 0.5$$

7. Arrange these magnitudes in descending order.

$$3 > 1 > 0.5 > 0 \;\Rightarrow\; F_{BC} > F_{AB} > F_{DE} > F_{CD}$$

Hence, the ranking from high to low is $F_{BC}>F_{AB}>F_{DE}>F_{CD}$, so the answer is option C.

1. Velocity is the rate of change of displacement with time, found by differentiating $x(t)$.

$$v(t) = \frac{dx}{dt} = 2\alpha t + \beta$$

2. Substitute $\alpha=1\text{ m/s}^2$ and $\beta=1\text{ m/s}$.

$$v(t) = 2t+1$$

3. Find the velocity at the start and end of the interval.

$$v_i = v(2) = 2(2)+1 = 5\text{ m/s}, \qquad v_f = v(3) = 2(3)+1 = 7\text{ m/s}$$

4. By the work-energy theorem, the work done equals the change in kinetic energy.

$$W = \frac{1}{2}m(v_f^2-v_i^2)$$

5. Substitute $m=2\text{ kg}$.

$$W = \frac{1}{2}(2)(7^2-5^2) = 49-25 = 24\text{ J}$$

Hence, the work done on the body is 24 J, so the answer is option B.

1. The general definition of work done by a force $\vec F$ while moving an object from $\vec r_1$ to $\vec r_2$ is the line integral of the force along the path, without any extra negative sign.

$$W = \int\limits_{r_1}^{r_2}\vec F\cdot d\vec r$$

2. This formula applies to both conservative and non-conservative forces. Statement I instead claims $W=-\int_{r_1}^{r_2}\vec F\cdot d\vec r$, which is incorrect, so Statement I is false.
3. For a conservative force specifically, the work done equals the negative change in potential energy, and this value depends only on the initial and final positions.

$$W = \int\limits_{\vec r_1}^{\vec r_2}\vec F\cdot d\vec r = -(U_f-U_i)$$

4. Since a conservative force satisfies $\oint\vec F\cdot d\vec r=0$ (zero net work over any closed loop), the work between two fixed points is path-independent.
5. Statement II claims the work done changes with the path taken for a conservative force — this directly contradicts path-independence, so Statement II is also false. Hence, both Statement I and Statement II are false, so the answer is option C.