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

The NCERT text states: 'If the displacement $\Delta x$ is small, we can take the force $F(x)$ as approximately constant and the work done is then $\Delta W = F(x) \Delta x$.'

As shown in the NCERT derivation for the work-energy theorem for a variable force: $\frac{dK}{dt} = \frac{d}{dt} (\frac{1}{2}mv^2) = mv \frac{dv}{dt}$. Since $m \frac{dv}{dt} = F$ (from Newton's Second Law), then $\frac{dK}{dt} = Fv$.

The NCERT text explicitly states: 'Thus, for a varying force the work done can be expressed as a definite integral of force over displacement: $W = \int_{x_i}^{x_f} F(x) dx$.'

For a varying force, the work done is represented by the area under the force-displacement curve. If the curve forms a triangle, calculating the area of the triangle($1/2 \times base \times height$) would yield the work done.

The work-energy theorem states that the net work done on an object equals the change in its kinetic energy ($\Delta K = W_{net}$). This holds true for both constant and variable forces, as explained under 'THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE'.

Given $m = 0.5 \text{ kg}$, $v = ax^{3/2}$, $a = 5 \text{ m}^{-1/2} \text{ s}^{-1}$. Initial velocity at $x=0$, $v_i = a(0)^{3/2} = 0$. Final velocity at $x=2 \text{ m}$, $v_f = a(2)^{3/2} = 5 \times (2^{3/2}) = 5 \times (2 \sqrt{2}) = 10 \sqrt{2} \text{ m/s}$. Work done by net force (Work-Energy Theorem) is $W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$. $W = \frac{1}{2} (0.5 \text{ kg}) (10 \sqrt{2} \text{ m/s})^2 - 0$ $W = \frac{1}{2} (0.5) (100 \times 2) = \frac{1}{2} (0.5) (200) = 0.5 \times 100 = 50 \text{ J}$. NOTE: Re-calculating. The given answer for similar problem 5.20 results in 125 J. Let's recheck the calculation of $v_f$: $v_f = 5 \times (2^{3/2}) = 5 \times (2 \cdot \sqrt{2}) = 10\sqrt{2}$. Then $v_f^2 = (10\sqrt{2})^2 = 100 \times 2 = 200$. So, $W = \frac{1}{2} \times 0.5 \times 200 = 50 \text{ J}$. Let's assume there's a misunderstanding of the problem from the textbook. The physics is about applying the work-energy theorem. Given the exact problem from NCERT (5.20), let's ensure the calculation is accurate. $v = 5 x^{3/2}$ $v_i = 0$ at $x=0$ $v_f = 5 (2)^{3/2} = 5 \times 2 \sqrt{2} = 10 \sqrt{2} \text{ m/s}$ at $x=2 \text{ m}$ $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} (0.5) (10\sqrt{2})^2 = \frac{1}{2} (0.5) (100 \times 2) = \frac{1}{2} (0.5) (200) = 50 \text{ J}$ $W = K_f - K_i = 50 - 0 = 50 \text{ J}$. However, if we are to derive the given solution from NCERT (which yields 125 J in the solution part of text related to similar problems), there must be a mismatch somewhere. Let's re-read the context. Ah, wait, this problem is actually part of the 'Additional Exercises' (Question 5.20) in the NCERT, for which the solution is not explicitly provided in the excerpt. My calculation gives 50 J. Let me ensure if there was any mistake in my understanding of the problem that could lead to 125 J. No, the calculation follows the work-energy theorem correctly. So 50 J is the correct value. Since it's an MCQ, let's assume the options are based on possible values, and the calculation of 50J is solid. Let's re-evaluate in case the question was implicitly asking for the work done by a force $F = ma = m \frac{dv}{dt}$. $v = ax^{3/2} \implies \frac{dv}{dt} = \frac{d}{dt} (ax^{3/2}) = a \frac{3}{2} x^{1/2} \frac{dx}{dt} = a \frac{3}{2} x^{1/2} v = a \frac{3}{2} x^{1/2} (ax^{3/2}) = \frac{3}{2} a^2 x^2$ $F = m \frac{dv}{dt} = m \frac{3}{2} a^2 x^2$ $W = \int F dx = \int_0^2 m \frac{3}{2} a^2 x^2 dx = m \frac{3}{2} a^2 \left[\frac{x^3}{3}\right]_0^2 = m \frac{3}{2} a^2 \frac{8}{3} = 4 m a^2$ Substitute values: $m = 0.5 \text{ kg}$, $a = 5 \text{ m}^{-1/2} \text{ s}^{-1}$ $W = 4 \times 0.5 \times (5)^2 = 2 \times 25 = 50 \text{ J}$. Both methods yield 50 J. So, the correct option should reflect 50 J. If 125 J was expected, the 'a' or velocity function might be different implicitly. Sticking to my calculation from the problem statement, 50 J is correct.

$U\propto x^2$; $(8/2)^2 = 16$, so $U_2 = 16U$.

$v = dx/dt = 2$ m/s; $P = Fv = 10$ W.

$mv_1 = 2m v_2 \Rightarrow v_1 : v_2 = 2 : 1$.