# A body of mass 5kg is moving with a momentum of 10 kgm/s. A force of 0.2N acts on it in the direction of motion of the body for 10 sec. The increase in its kinetic energy is?

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#### Explanation

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#### Explanation:

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Dec 13, 2016

$\text{4.4 J}$

#### Explanation:

Before the force is applied we know

$\text{momentum" = "mass" times "velocity" = "10 kg m/s}$

Since mass in the above is $\text{5 kg}$, then velocity = ${\text{2 ms}}^{- 1}$

We can now work out kinetic energy, KE:

$K E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot 5 \cdot {2}^{2} = \text{10 J}$

The force of $\text{0.2 N}$ is now applied in the direction of travel for $10$ seconds. Newton's second law can be expressed as "the force applied equals the rate of change of momentum". OR

$F = \frac{m v - \text{mu}}{t} = m \cdot \frac{v - u}{t}$

Where $v$ is the final velocity after applying force, and $u$ is the initial velocity ( ${\text{2 ms}}^{- 1}$ in this case from above)

So

$0.2 = 5 \cdot \frac{\left(v - 2\right)}{10}$

rearranging helps us find $v$

$2 = 5 \cdot \left(v - 2\right)$

$\frac{2}{5} = v - 2$

$v = \frac{12}{5} {\text{ ms}}^{- 1}$

Hence final kinetic energy, $K E$, is

$K E = \frac{1}{2} m {v}^{2} = \frac{1}{2} \cdot 5 \cdot {\left(\frac{12}{5}\right)}^{2} = \text{14.4 J}$

So the overall change in $K E$ is

${\Delta}_{\text{KE" = 14.4 - 10 = "4.4 J}}$

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