Work Energy and Power - Result Question 2
2. An object of mass $500 g$, initially at rest, is acted upon by a variable force whose Xcomponent varies with $X$ in the manner shown. The velocities of the object at the points $X=8$ $m$ and $X=12 m$, would have the respective values of (nearly)
[NEET Odisha 2019]
(a) $18 m / s$ and $20.6 m / s$
(b) $18 m / s$ and $24.4 m / s$
(c) $23 m / s$ and $24.4 m / s$
(d) $23 m / s$ and $20.6 m / s$
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Answer:
Correct Answer: 2. (d)
Solution:
- (d) Using work-energy theorem, $\Delta K=$ work $=$ area under $F-x$ graph
From $x=0$ to $x=8 m$
$\frac{1}{2} m v^{2}=100+30$
$\therefore v=\sqrt{520}=23 m / s$
$\frac{W_p}{W_Q}=\frac{\frac{1}{2} k_p x_p^{2}}{\frac{1}{2} k_Q x_Q^{2}}=\frac{k_p \cdot x_p \cdot x_p}{k_Q \cdot x_Q \cdot x_Q}=\frac{F x_p}{F x_Q}=\frac{k_Q}{k_p}$
As $k_P>k_Q \quad \therefore \quad W_Q>W_P$
