Question #1c2bd

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Question #1c2bd

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Answer 1 · 2016-04-27T08:47:37

$4.4 J$ $"4.4 J"$

Explanation:

Before the force is applied we know

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

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

We can now work out kinetic energy, KE:

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \cdot 5 \cdot 2^{2} = 10 J$ $KE=1/2mv^2=1/2*5*2^2="10 J"$

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

$F = \frac{m v - mu}{t} = m \cdot \frac{v - u}{t}$ $F=(mv-"mu")/t=m * (v-u)/t$

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

So

$0.2 = 5 \cdot \frac{(v - 2)}{10}$ $0.2 = 5 * ((v-2))/10$

rearranging helps us find $v$ $v$

$2 = 5 \cdot (v - 2)$ $2 = 5 * (v-2)$

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

$v = \frac{12}{5} {ms}^{- 1}$ $v = 12/5" ms"^(-1)$

Hence final kinetic energy, $K E$ $KE$ , is

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \cdot 5 \cdot {(\frac{12}{5})}^{2} = 14.4 J$ $KE = 1/2mv^2 = 1/2 * 5 * (12/5)^2 = "14.4 J"$

So the overall change in $K E$ $KE$ is

$Delta_"KE" = 14.4 - 10 = "4.4 J"$

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Question #1c2bd

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