Work Power and Energy 1 Question 2

3. A force acts on a $2 kg$ object, so that its position is given as a function of time as $x=3 t^{2}+5$. What is the work done by this force in first 5 seconds?

(a) $850 J$

(b) $900 J$

(c) $950 J$

(d) $875 J$

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Answer:

Correct Answer: 3. (b)

Solution:

  1. Here, the displacement of an object is given by

$$ x=\left(3 t^{2}+5\right) m $$

Therefore, velocity $(v)=\frac{d x}{d t}=\frac{d\left(3 t^{2}+5\right)}{d t}$

or

$$ v=6 t m / s $$

The work done in moving the object from $t=0$ to $t=5 s$

$$ W=\int _{x _0}^{x _5} F \cdot d x $$

The force acting on this object is given by

$$ \begin{aligned} F & =m a=m \times \frac{d v}{d t} \\ & =m \times \frac{d(6 t)}{d t} \quad[\therefore \text { using (i) }] \\ F & =m \times 6=6 m=12 N \end{aligned} $$

Also, $\quad x _0=3 t^{2}+5=3 \times(0)^{2}+5=5 m$

and at $t=5 s$,

$$ x _5=3 \times(5)^{2}+5=80 m $$

Put the values in Eq. (ii),

$$ W=12 \times \int _{x _0}^{x _5} d x=12[80-5] $$

$$ W=12 \times 75=900 J $$

Alternative Method

To using work - kinetic energy theorem is,

$$ \begin{aligned} W & =\Delta K \cdot E=\frac{1}{2} m\left(v _f^{2}-v _i^{2}\right) \\ & =\frac{1}{2} m \times\left(30^{2}-0^{2}\right)=\frac{1}{2} \times 2 \times 900=900 J \end{aligned} $$



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