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 $$\Delta K=0$$.
Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
According to work-energy theorem,
Change in $$\mathrm{KE}=$$ Work done by the retarding force
KE of the body $$=$$ Retarding force $\times$ Displacement
As KE of the bodies and retarding forces applied on them are same, therefore, both bodies will travel equal distances before coming to rest.
A bob of mass $$m$$ suspended by a light string of length $$L$$ is whirled into a vertical circle as shown in figure. What will be the trajectory of the particle, if the string is cut at
(a) point $$B$$ ?
(b) point $$C$$ ?
(c) point $$X$$ ?
When bob is whirled into a vertical circle, the required centripetal force is obtained from the tension in the string. When string is cut, tension in string becomes zero and centripetal force is not provided, hence, bob start to move in a straight line path along the direction of its velocity.
(a) At point $$B$$, the velocity of $$B$$ is vertically downward, therefore, when string is cut at $$B$$, bob moves vertically downward.
(b) At point $$C$$, the velocity is along the horizontal towards right, therefore, when string is cut at $$C$$, bob moves horizontally towards right.
Also, the bob moves under gravity simultaneously with horizontal uniform speed. So, it traversed on a parabolic path with vertex at $$C$$.
(c) At point $$X$$, the velocity of the bob is along the tangent drawn at point $$X$$, therefore when string is cut at point $$C$$, bob moves along the tangent at that point $$X$$.
Also, the bob move under gravity simultaneously with horizontal uniform speed. So, it traversed on a parabolic path with vertex higher than $$C$$.
A graph of potential energy $$V(x)$$ versus $$x$$ is shown in figure. A particle of energy $$E_0$$ is executing motion in it. Draw graph of velocity and kinetic energy versus $$x$$ for one complete cycle $$A F A$$.
$$\begin{aligned} &\text { KE versus } \boldsymbol{x} \text { graph }\\ &\begin{array}{lrl} \text { We know that } & \text { Total } \mathrm{ME} & =\mathrm{KE}+\mathrm{PE} \\ \Rightarrow & E_0 & =\mathrm{KE}+\mathrm{V}(x) \\ \Rightarrow & \mathrm{KE} & =E_0-V(x) \end{array} \end{aligned}$$
$$\text { at } A_1 x=0, V(x)=E_0$$
$$\Rightarrow \quad \mathrm{KE}=E_0-E_0=0$$
$$ \begin{aligned} & \text { at } B_1 V(x)< E_0 \\ & \Rightarrow \quad \text { KE } > 0 \quad \text{(positive)} \end{aligned}$$
at C and $$D_1 V(x)=0$$
$$\Rightarrow \mathrm{KE}$$ is maximum at $$F_1 V(x)=E_0$$
Hence, $$\mathrm{KE}=0$$
The variation is shown in adjacent diagram.
Velocity versus $$x$$ graph
As $$\mathrm{KE}=\frac{1}{2} m v^2$$
$$\therefore$$ At $$A$$ and $$F$$, where $$K E=0, v=0$$.
At $$C$$ and $$D, \mathrm{KE}$$ is maximum. Therefore, $$v$$ is $$\pm \max$$.
At $$B, \mathrm{KE}$$ is positive but not maximum.
Therefore, $$\quad v$$ is $$\pm$$ some value $$\quad$$ (< max.)
The variation is shown in the diagram.
