Work Energy and Power - Result Question 19
21. A body of mass $1 kg$ is thrown upwards with a velocity $20 m / s$. It momentarily comes to rest after attaining a height of $18 m$. How much energy is lost due to air friction? $(g=10$ $m / s^{2}$ )
[2009]
(a) $30 J$
(b) $40 J$
(c) $10 J$
(d) $20 J$
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Answer:
Correct Answer: 21. (d)
Solution:
- (d) When the body is thrown upwards. its K.E is converted into P.E.
The loss of energy due to air friction is the difference of K.E and P.E.
$ \begin{aligned} \frac{1}{2} mv^{2}-mgh= & \frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \\ & =200-180=20 J \end{aligned} $
22 .
(b) $E=\frac{p^{2}}{2 m}$
or, $E_1=\frac{p_1^{2}}{2 m_1}, E_2=\frac{p_2^{2}}{2 m_2}$
or, $m_1=\frac{p_1^{2}}{2 E_1}, m_2=\frac{p_2^{2}}{2 E_2}$
$m_1>m_2 \Rightarrow \frac{m_1}{m_2}>1$
$\therefore \frac{p_1^{2} E_2}{E_1 P_2^{2}}>1 \Rightarrow \frac{E_2}{E_1}>1 \quad[\because p_1=p_2]$
or, $E_2>E_1$
Kinetic energy of body of mass $m$ moving with velocity $v$ is
$ \text{ K.E. }=\frac{1}{2} m v^{2} $
$\Rightarrow$ K.E. $=\frac{1}{2} \frac{m v^{2} \times m}{m}$
[Multiplying both numerator and denominator by $m$ ]
$\Rightarrow$ K.E. $=\frac{1}{2} \frac{m^{2} v^{2}}{m}$
$\Rightarrow$ K.E. $=\frac{1}{2} \frac{p^{2}}{m} \quad[\because$ Momentum, $P=m v]$
$\Rightarrow P^{2}=2 m$ K.E.
$\Rightarrow P=\sqrt{2 m K . E}$.
