A mass \(m\) is attached to a string and moves in a vertical circle of constant radius \(R\). At the top of the circle, the tension in the string is \(T\). Air resistance is negligible.
What is the kinetic energy of \(m\) at the top of the circle?
Correct Answer: \( \boxed{\mathrm{B}} \)
At the top of the vertical circle, both the tension \(T\) and the weight \(mg\) act towards the centre of the circle.
The resultant centripetal force is therefore
\(T+mg=\dfrac{mv^2}{R}\)
Rearranging,
\(mv^2=R(T+mg)\)
The kinetic energy is
\(K=\dfrac{1}{2}mv^2\)
Therefore,
\(K=\dfrac{R(T+mg)}{2}\)
Thus, the kinetic energy of the mass at the top of the circle is
\( \boxed{\dfrac{R(T+mg)}{2}} \)
Hence, the correct answer is \( \boxed{\mathrm{B}} \).
A pump has efficiency \(\eta\) when raising water from a well of depth \(d\). The mass of water raised per second is \(R\). Changes in kinetic energy of the water are considered negligible.
What is the input power to the pump required to raise the water?
Correct Answer: \( \boxed{\mathrm{D}} \)
The mass of water raised per second is \(R\), so the useful mass flow rate is
\(\dfrac{m}{t}=R\)
The gravitational potential energy gained per unit time is therefore
\(P_{\mathrm{useful}}=Rgd\)
The efficiency of the pump is
\(\eta=\dfrac{P_{\mathrm{useful}}}{P_{\mathrm{input}}}\)
Therefore,
\(P_{\mathrm{input}}=\dfrac{P_{\mathrm{useful}}}{\eta}\)
\(P_{\mathrm{input}}=\dfrac{Rgd}{\eta}\)
Thus, the required input power is
\( \boxed{\dfrac{Rgd}{\eta}} \)
Hence, the correct answer is \( \boxed{\mathrm{D}} \).
A block of mass \(m\) is released from rest and slides down a ramp of length \(d\) that makes an angle \(\theta\) with the horizontal. A constant frictional force \(F\) acts on the block.
What is the kinetic energy of the block at the bottom of the ramp?
Correct Answer: \( \boxed{\mathrm{D}} \)
The block starts from rest, so its initial kinetic energy is zero. Using the work-energy theorem, the final kinetic energy equals the net work done on the block.
The vertical height through which the block falls is
\(h=d\sin\theta\)
Therefore, the work done by gravity is
\(W_g=mgh=mgd\sin\theta\)
The frictional force acts opposite to the displacement, so its work is
\(W_F=-Fd\)
Hence, the net work is
\(W_{\mathrm{net}}=mgd\sin\theta-Fd\)
Since the initial kinetic energy is zero,
\(K_{\mathrm{final}}=W_{\mathrm{net}}\)
Therefore,
\( \boxed{K=mgd\sin\theta-Fd} \)
Hence, the correct answer is \( \boxed{\mathrm{D}} \).
