Work Energy and Power Questions and Answers

Answer: B

The kinetic energy is proportional to the square of the speed. If the bullet keeps 20 percent of its kinetic energy, then \(v^2 = 0.2 \times 100^2 = 2000\), so

The mass and the thickness of the plywood are not needed. A common slip is to take 20 percent of the speed, giving 20 m/s, or to use 0.2 times the energy without the square root; none of the other options, 34.70, 38.78 and 40.74 m/s, corresponds to any fraction of the energy that the question gives.

Answer: B

The work done is against gravity over the total height climbed. The height is 40 steps x 0.40 m = 16 m, so the work is \(mgh = 60 \times 10 \times 16 = 9600\) J. The power is the work divided by the time:

The value 240 W would come from halving the height, and 720 W and 960 W from using the wrong time or a height of 24 m or 32 m; the data give exactly 16 m and 20 s.

Answer: B

The kinetic energy is about 790 J.

A solid disc has \(I = \tfrac{1}{2}mr^{2}\):

\[ \begin{aligned} I &= \tfrac{1}{2}(20)(0.2)^{2} = 0.4\ \text{kg m}^{2} \\ \omega &= \frac{2\pi(600)}{60} = 62.83\ \text{rad/s} \\ KE &= \tfrac{1}{2}(0.4)(62.83)^{2} = 789.6\ \text{J} \end{aligned} \]

1580 J uses \(I = mr^{2}\), the value for a thin ring, and 3160 J also drops the half in the kinetic energy formula. 395 J is half the correct value. The thickness given is not needed once the mass is known.

Answer: C

The ram should be released from 35.1°.

Because the two ropes are equal and parallel, the ram does not rotate: it moves in curvilinear translation, every point travelling on a circle of radius 7 m. Its size (9 m by 2 m) therefore does not enter. When released from angle \(\theta\), its centre of mass falls by \(7(1 - \cos\theta)\), and energy conservation gives the speed at the bottom:

\[ \begin{aligned} v^{2} &= 2gL(1 - \cos\theta) \\ 1 - \cos\theta &= \frac{25}{2 \times 9.81 \times 7} = 0.1820 \\ \cos\theta &= 0.8180 \Rightarrow \theta = 35.1^{\circ} \end{aligned} \]

The larger angles would give striking speeds well above 5 m/s; for example 67.1° corresponds to about 9.2 m/s.