Work, Energy and Power — NEET UG practice

1. Since the crane lifts the mass against gravity, the work done equals the gain in potential energy, and power is work done divided by time taken.

$P=\frac{W}{t}=\frac{mgh}{t}$

2. Substitute the given values $m=1000\text{ kg}$, $g=9.8\text{ m/s}^2$, $h=20\text{ m}$, $t=10\text{ s}$.

$P=\frac{1000\times9.8\times20}{10}$

3. Simplifying this expression.

$P=19600\text{ W}=19.6\text{ kW}$

Hence, the answer is C.

1. Since the friction coefficient becomes zero exactly at $x=L$, use this condition to relate $\mu_0$ and $\alpha$.

$\mu_k(L)=0 \;\Rightarrow\; \mu_0=\alpha L$

2. On a horizontal surface the normal force equals the weight, so the frictional force is $f_k=\mu_k Mg$. The work done by friction is found by integrating this force over the path from $x=0$ to $x=L$.

$W_f=\int_0^{L}(\mu_0-\alpha x)\,Mg\,dx$

3. Carrying out the integration term by term.

$W_f=\mu_0 MgL-\alpha Mg\frac{L^2}{2}$

4. Substitute $\alpha=\dfrac{\mu_0}{L}$ from Step 1 into this result.

$W_f=\mu_0 MgL-\frac{\mu_0}{L}\cdot Mg\cdot\frac{L^2}{2}=\frac{\mu_0 MgL}{2}$

5. Comparing this with the given expression $n\mu_0 MgL$.

$n=\frac{1}{2}$

Hence, the answer is A.

1. By the work-energy theorem, the work done by the braking force equals the loss in kinetic energy. Since both cars come to a complete stop, their final kinetic energy is zero.

$F\cdot S=K_i$

2. Apply this relation separately for car $A$ and car $B$.

$F_A S_A=K_A,\qquad F_B S_B=K_B$

3. Dividing these two equations gives the ratio of the braking forces.

$\frac{F_A}{F_B}=\frac{K_A}{K_B}\times\frac{S_B}{S_A}$

4. Substitute $K_A=100\text{ J}$, $K_B=225\text{ J}$, $S_A=1000\text{ m}$, $S_B=1500\text{ m}$.

$\frac{F_A}{F_B}=\frac{100}{225}\times\frac{1500}{1000}$

5. Simplifying the fraction.

$\frac{F_A}{F_B}=\frac{2}{3}$

Hence, the answer is D.