A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.
No, work done in moving along a closed loop is not necessarily zero. It is zero only when all the forces are conservative forces.
In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact)?
(a) Kinetic energy.
(b) Total linear momentum.
Give reason for your answer in each case.
Total linear momentum of the system of two balls is always conserved. While balls are in contact, there may be deformation which means elastic PE which came from part of KE Therefore, KE may not be conserved.
Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.
$$\begin{aligned} \text { mass } & =m=100 \mathrm{~kg} \\ \text { height } & =h=10 \mathrm{~m} \text { time duration } t=20 \mathrm{~s} \\ \text { power } & =\text { Rate of work done } \\ & =\frac{\text { change of } \mathrm{PE}}{\text { time }}=\frac{m g h}{t} \\ & =\frac{100 \times 9.8 \times 10}{20} \\ & =5 \times 98=490 \mathrm{~W} \end{aligned}$$
The average work done by a human heart while it beats once is 0.5 J . Calculate the power used by heart if it beats 72 times in a minute.
Given, average work done by a human heart per beat $$=0.5 \mathrm{~J}$$
$$\begin{aligned} &\text { Total work done during } 72 \text { beats }\\ &\begin{aligned} & =72 \times 0.5 \mathrm{~J}=36 \mathrm{~J} \\ \text { Power } & =\frac{\text { Work done }}{\text { Time }}=\frac{36 \mathrm{~J}}{60 \mathrm{~s}}=0.6 \mathrm{~W} \end{aligned} \end{aligned}$$
Give example of a situation in which an applied force does not result in a change in kinetic energy.
When a charged particle moves in a uniform normal magnetic field, the path of the particle is circular, as given field is uniform hence, radius of the circular path is also constant.
As the force is central and movement is tangential work done by the force is zero. As speed is also constant we can say that $$\Delta K=0$$.