Question 10: Work, Energy and Power & Circular Motion

Question

A body of mass 2 kg is moving along x-direction such that its displacement as function of time is given by $x(t) = \alpha t^2 + \beta t + \gamma$ m, where $\alpha = 1 \; m/s^2$, $\beta = 1 \; m/s$ and $\gamma = 1 \; m$. The work done on the body during the time interval $t = 2 \; s$ to $t = 3 \; s$, is ________ J.

Accepted Answer

Velocity ( $v$ ) is the rate of change of displacement ( $x$ ) with respect to time ( $t$ ) :
$$ v(t)=\frac{d x}{d t} $$
The displacement function is given as:
$$ x(t)=\alpha t^2+\beta t+\gamma $$
Differentiating with respect to $t$ :
$$ v(t)=\frac{d}{d t}\left(\alpha t^2+\beta t+\gammaight) $$
$$\Rightarrow $$ $$ v(t)=2 \alpha t+\beta $$
Substituting the given values $\left(\alpha=1 \mathrm{~m} / \mathrm{s}^2, \beta=1 \mathrm{~m} / \mathrm{s}ight)$ :
$$ v(t)=2(1) t+1 $$
$$\Rightarrow $$ $$ v(t)=2 t+1 $$
Initial velocity $\left(\mathrm{v}_{\mathrm{i}}ight)$ at $\mathrm{t}=2 \mathrm{~s}$ :
$$ \mathrm{v}_{\mathrm{i}}=2(2)+1=5 \mathrm{~m} / \mathrm{s} $$
Final velocity $\left(\mathrm{v}_{\mathrm{f}}ight)$ at $\mathrm{t}=3 \mathrm{~s}$ :
$$ v_f=2(3)+1=7 \mathrm{~m} / \mathrm{s} $$
Using Work-Energy Theorem formula,
$\mathrm{W}=\Delta \mathrm{K} \cdot \mathrm{E} \cdot=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{f}}^2-\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{i}}^2$
$$\Rightarrow $$ $\mathrm{W}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{f}}^2-\mathrm{v}_{\mathrm{i}}^2ight)$
$$\Rightarrow $$ $\mathrm{W}=\frac{1}{2} 	imes 2 	imes\left(7^2-5^2ight)$
$$\Rightarrow $$ $\mathrm{W}=1 	imes(49-25)$
$$\Rightarrow $$ $$ \mathrm{W}=24 \mathrm{~J} $$
Therefore, the work done on the body during the given time interval is 24 J .
Hence, the correct option is (B).

Answer

42

Answer

24

Answer

49

Answer

12

Work, Energy and Power & Circular Motion

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