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) $875 J$
Show Answer
Answer:
Correct Answer: 3. (b)
Solution:
- 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,
$$ \begin{aligned} W & =\Delta K \cdot E=\frac{1}{2} m\left(v _f^{2}-v _i^{2}\right) \\ & =\frac{1}{2} m \times\left(30^{2}-0^{2}\right)=\frac{1}{2} \times 2 \times 900=900 J \end{aligned} $$
