# The kinetic energy of an object with a mass of 5 kg constantly changes from 72 J to 480 J over 12 s. What is the impulse on the object at 2 s?

Apr 19, 2016

#### Answer:

Assume that the kinetic energy is increasing at a constant rate. After 2s, the impulse on the object would have been $10.58 \setminus \quad K g \setminus \cdot \frac{m}{s}$

#### Explanation:

The impulse exerted on an object equals the change it in its momentum

$I m p = \setminus \Delta p = m \left({v}_{f} - {v}_{i}\right)$

The object's initial kinetic energy is 72 J, so
$72 J = \frac{1}{2} m {v}_{i}^{2} \setminus \quad \setminus \quad \setminus \implies {v}_{i} = 5.37 \frac{m}{s}$

To find the impulse on the object at 2s we need to find the speed of the object, ${v}_{f}$, at 2s.

We are told that the kinetic energy changes constantly. The kinetic energy changes by $\left(480 J - 72 J = 408 J\right)$ over 12 seconds.

This means that the kinetic energy changes at a rate of:
$\frac{408 J}{12 s} = 34 \frac{J}{s}$

In two seconds the kinetic energy will have increased by $34 \frac{J}{s} \cdot 2 s = 68 J$

Therefore, at 2s the kinetic energy is $\left(72 J + 68 J\right) = 140 J$. This allows us to solve for the ${v}_{f}$ at 2s

$140 J = \frac{1}{2} m {v}_{f}^{2} \setminus \quad \setminus \quad \setminus \implies {v}_{f} = 7.48 \frac{m}{s}$

Now we have to make sure ${v}_{f}$ and ${v}_{i}$ have the right signs when we find $\setminus \Delta p$. Assuming the kinetic energy is constantly increasing, ${v}_{f}$ and ${v}_{i}$ will be in the same direction and have the same sign.

Substitute $m$, ${v}_{i}$, and ${v}_{f}$ to solve for the impulse.
$I m p = \setminus \Delta p = \left(5 K g\right) \left(7.48 \frac{m}{s} - 5.37 \frac{m}{s}\right) = 10.58 \setminus \quad K g \setminus \cdot \frac{m}{s}$