# Finding Work Done, Acceleration, Force and Time

1. Apr 16, 2013

1. The problem statement, all variables and given/known data
An object with a mass of 2kg gets accelerated from 15m/s to 25m/s.

A: What is the work done in effecting this velocity change?
B: The distance moved was 8m, what force produced the acceleration?
C: How long did it take to produce the acceleration (presumed uniform)?

2. Relevant equations
Equations of Motion

3. The attempt at a solution
Part A:
$$W=\frac{1}{2}m(v^2 - u^2) \\ W=\frac{1}{2}2(25^2 - 15^2) \\ W= 25^2 - 15^2 = 400J$$

Part B:
$$v^2=u^2+2as \\ ∴ a = \frac{v^2 - u^2}{2s} \\ F=ma \\ F=2 \times \frac{25^2 - 15^2}{2 \times 8} \\ F=2 \times \frac{400}{16} \\ F=2 \times 25 = 50N$$

Part C
$$v=u+at \\ ∴ t = \frac{v-u}{a} \\ t= \frac{25-15}{25} = 0.4s$$

My biggest problem is the values, they just dont seem right to me, 0.4s to accelerate a 2kg mass from 15 to 25m/s using 50N over 8m?

Any help appreciated :).

2. Apr 16, 2013

### Clever_name

look at the definition of work, for part b.

3. Apr 16, 2013

### ap123

For part (c), you can use the impulse-momentum theorem to check your answer

4. Apr 16, 2013

I keep reading it but cant seem to find another way of interpreting it. Surely it means what force produced the acceleration (which is 25m/s^2 from my calculations) and then that comes out to be 50N right?

Any little hint at just interpreting it would be appreciated :) .

5. Apr 16, 2013

Are right, is that that the impulse is equal to the change in momentum?

that would be then
$$I=Ft \\ I=50 \times 0.4 = 20Ns \\ Δp=mv-mu \\ Δp=2 \times 25 - 2 \times 15 = 50-30 = 20Ns$$

Does that mean it is correct?

6. Apr 16, 2013

### ap123

It looks right :)
It's always useful if you can find another way of doing a problem to check your answers.

For part (b) you can do a similar answer check by using the definition of work (as mentioned by Clever_name).

7. Apr 16, 2013