Question 3: Work, Energy and Power & Circular Motion

Question

The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6 m above the ground is $\_\_\_\_$ $\mathrm{m} / \mathrm{s}$. Assume pulley is massless and string is light and inextensible. (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

For the 6 kg mass $(\mathrm{M})$ moving down and the 2 kg mass $(\mathrm{m})$ moving up, let the acceleration of system is a and tension in the string is T .
Using Newton's second law ( $\mathrm{F}=\mathrm{ma}$ ) of motion for M
$$\mathrm{Mg}-\mathrm{T}=\mathrm{Ma} \ldots (i) $$
For m ,
$$ \mathrm{T}-\mathrm{mg}=\mathrm{ma} \ldots (ii)$$
Adding the two equations :
$$ (M-m) g=(M+m) a $$
$$\Rightarrow $$ $$ a=\frac{(M-m)}{M+m} g $$
Substituting $\mathrm{M}=6, \mathrm{~m}=2$, and $\mathrm{g}=10$ :
$$ \mathrm{a}=\frac{6-2}{6+2} 	imes 10=\frac{4}{8} 	imes 10=5 \mathrm{~m} / \mathrm{s}^2 $$
Using the kinematic equation for constant acceleration, starting from rest ( $u=0$ ) :
$$ \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as} $$
$$\Rightarrow $$ $$ \mathrm{v}=\sqrt{2 \mathrm{as}} $$
Where s is the height $\mathrm{h}=6 \mathrm{~m}$.
$$ v=\sqrt{2 	imes 5 	imes 6} $$
$$\Rightarrow $$ $$ \mathrm{v}=\sqrt{60} \approx 7.745 \mathrm{~m} / \mathrm{s} $$
Therefore, the velocity is approximately $7.74 \mathrm{~m} / \mathrm{s}$.
The correct option is (A).

Answer

7.74

Answer

7.20

Answer

6.55

Answer

4.50

Work, Energy and Power & Circular Motion

The velocity at which 6 kg mass (shown in figure) strikes the ground when it is released from a height of 6 m above the ground is \_\_\_\_ \mathrm{m} / \mathrm{s}. Assume pulley is massless and string is light and inextensible. (Take \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2 )

Choose the correct answer: