Question 5: Work, Energy and Power & Circular Motion

Question

The rain drop of mass 1 g , starts with zero velocity from a height of 1 km . It hits the ground with a speed of $5 \mathrm{~m} / \mathrm{s}$. The work done by the unknown resistive force is $\_\_\_\_$ J.
(take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

The Work-Energy Theorem states that the total work done by all forces (conservative and non-conservative) on a particle is equal to the change in its kinetic energy.
$$ \mathrm{W}_{	ext {total }}=\Delta \mathrm{K} $$
When rain drop is falling, it experiences gravitational force (vertically downward) and resistive force which opposes motion of rain drop (vertically upward).
$$ \mathrm{W}_{	ext {gravity }}+\mathrm{W}_{	ext {resistive }}=\mathrm{K}_{	ext {final }}-\mathrm{K}_{	ext {initial }} $$
Work done by gravity when an object falls a height h is:
$$ \mathrm{W}_{\mathrm{g}}=\mathrm{F}_{\mathrm{g}} \mathrm{~h}=\mathrm{mgh} $$
The mass of the rain drop is $\mathrm{m}=1 \mathrm{~g}=1 	imes 10^{-3} \mathrm{~kg}$
The height of drop is $\mathrm{h}=1 \mathrm{~km}=1000 \mathrm{~m}$
And the acceleration due to gravity is $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
Putting the values,
$$ W_g=\left(10^{-3}ight) 	imes(10) 	imes(1000)=10 \mathrm{~J} $$
The initial velocity is $u=0$ as it started from rest, so the initial kinetic energy is $K_{	ext {initial }}=\frac{1}{2} m u^2=0$.
The final velocity when rain drop hits the ground is $\mathrm{v}=5 \mathrm{~m} / \mathrm{s}$.
Thus, the final kinetic energy of the rain drop is
$$ \mathrm{K}_{\mathrm{final}}=\frac{1}{2} \mathrm{mv}^2 $$
$$\Rightarrow $$ $K_{f i n a l}=\frac{1}{2} 	imes\left(10^{-3}ight) 	imes(5)^2$
$$\Rightarrow $$ $$ \mathrm{K}_{	ext {final }}=\frac{1}{2} 	imes 10^{-3} 	imes 25=12.5 	imes 10^{-3} \mathrm{~J}=0.0125 \mathrm{~J} $$
Using the work-energy theorem equation,
$$ \mathrm{W}_{\mathrm{g}}+\mathrm{W}_{\mathrm{r}}=\mathrm{K}_{\mathrm{final}}-\mathrm{K}_{\mathrm{initial}} $$
$$\Rightarrow $$ $10 \mathrm{~J}+\mathrm{W}_{\mathrm{r}}=0.0125 \mathrm{~J}$
$$\Rightarrow $$ $\mathrm{W}_{\mathrm{r}}=0.0125-10$
$$\Rightarrow $$ $$ \mathrm{W}_{\mathrm{r}}=-9.9875 \mathrm{~J} $$
Therefore, the magnitude of the work done by the resistive force is approximately 9.98 J .
Thus, the correct option is (D).

Answer

-8.75

Answer

-8.35

Answer

-9.55

Answer

-9.98

Work, Energy and Power & Circular Motion

The rain drop of mass 1 g , starts with zero velocity from a height of 1 km . It hits the ground with a speed of 5 \mathrm{~m} / \mathrm{s}. The work done by the unknown resistive force is \_\_\_\_ J. (take \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 )

Choose the correct answer: