Work Done by Forces: notes and previous year questions

Answer: Option A

Since $\mu_{k}(L) = 0$, $\alpha L = \mu_{0}$. The friction is $\mu_{k}Mg = (\mu_{0} - \alpha x)Mg$, and its size summed over the path is $\displaystyle\int_{0}^{L}(\mu_{0} - \alpha x)Mg\,dx = Mg\left(\mu_{0}L - \dfrac{\alpha L^{2}}{2}\right) = Mg\left(\mu_{0}L - \dfrac{\mu_{0}L}{2}\right) = \tfrac{1}{2}\mu_{0}MgL$. So $n = \tfrac{1}{2}$. On a graph it is simply a triangle: the friction falls in a straight line from $\mu_{0}Mg$ to zero over the length $L$. The work done BY friction is $-\tfrac{1}{2}\mu_{0}MgL$, since it opposes the motion; the question asks only for its size. Option C keeps $\mu$ at $\mu_{0}$ all the way, as though the floor never got smoother.

Answer: Option C

Undo each term: $\alpha$ gives $\alpha x$ and $\beta x^{2}$ gives $\dfrac{\beta x^{3}}{3}$. From $0$ to $1\ \mathrm{m}$ the work is $\alpha + \dfrac{\beta}{3} = 1 + \dfrac{\beta}{3}$, and setting that to $5$ gives $\dfrac{\beta}{3} = 4$, so $\beta = 12\ \mathrm{N/m^{2}}$. Option A, $15$, is what comes of forgetting the constant force's share and writing $\dfrac{\beta}{3} = 5$ — the steady $1\ \mathrm{N}$ pushes through the whole metre too, and its $1\ \mathrm{J}$ is part of the $5$.

Answer: 8 J

First the motion. $a = \dfrac{(0.72 - 0.36)g}{0.72 + 0.36} = \dfrac{g}{3} = \dfrac{10}{3}\ \mathrm{m/s^{2}}$, with the $0.36\ \mathrm{kg}$ block going up. For that block $T - 0.36g = 0.36a$, so $T = 0.36\left(10 + \dfrac{10}{3}\right) = 4.8\ \mathrm{N}$. In the first second it rises $s = \tfrac{1}{2} \times \dfrac{10}{3} \times 1^{2} = \dfrac{5}{3}\ \mathrm{m}$. The string pulls up on a block moving up, so $W = 4.8 \times \dfrac{5}{3} = 8\ \mathrm{J}$. The same tension does $-8\ \mathrm{J}$ on the heavier block, which moves down against it — a light string's works on its two ends always add to zero. Using $T = 0.36g$, as if the block were still at rest, gives $6\ \mathrm{J}$.

Answer: Options A, B, D

Rising through a height $h$, the ball moves against its weight: $W = mgh\cos 180^{\circ} = -mgh$. Falling back through the same $h$, it moves along its weight: $W = +mgh$. The two cancel, so over the round trip gravity's total work is zero — the ball ends where it began. C mistakes a constant force for a force that does no work; a constant force does work whenever its body moves along it. E mixes up work with position: at the top the ball is not moving, so at that instant gravity is doing no work at all.

Answer: 2 J

$G = x^{3} - 3x$, so $W = G(2) - G(0) = (8 - 6) - 0 = 2\ \mathrm{J}$. Split at $x = 1\ \mathrm{m}$ as a check: from $0$ to $1$ the work is $G(1) - G(0) = -2\ \mathrm{J}$, the force holding the body back; from $1$ to $2$ it is $G(2) - G(1) = 2 - (-2) = +4\ \mathrm{J}$. The net $+2\ \mathrm{J}$ is the same, and the single subtraction got there without splitting anything. Dropping the $-3$ gives $8\ \mathrm{J}$, as if the backward part of the force did nothing.

Answer: Options A, B, D

With a zero total, $W_{\mathrm{you}} + (-mgh) + 0 + W_{f} = 0$, and $W_{f}$ is negative, so $W_{\mathrm{you}} = mgh + |W_{f}|$: you pay for the height and for the friction. B is gravity ignoring the path, as always. C is false — the normal reaction does push, but at right angles to the slope the block moves along, so it does nothing. E is false as well: friction acts along the whole length of the slope, so a longer, gentler incline up to the same height costs more work against friction, not the same.