Finding Work Done, Acceleration, Force and Time

[FaraDazed]

Homework Statement

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)?

Homework Equations

Equations of Motion

The Attempt at a Solution

Part A:
<br /> W=\frac{1}{2}m(v^2 - u^2) \\<br /> W=\frac{1}{2}2(25^2 - 15^2) \\<br /> W= 25^2 - 15^2 = 400J<br />

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

Part C
<br /> v=u+at \\<br /> ∴ t = \frac{v-u}{a} \\<br /> t= \frac{25-15}{25} = 0.4s<br />

My biggest problem is the values, they just don't seem right to me, 0.4s to accelerate a 2kg mass from 15 to 25m/s using 50N over 8m?

Any help appreciated :).

 

[Clever_name]

look at the definition of work, for part b.

 

[ap123]

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

 

[FaraDazed]

look at the definition of work, for part b.

I keep reading it but can't 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 :) .

 

[FaraDazed]

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

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

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

Does that mean it is correct?

 

[ap123]

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

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

Does that mean it is correct?

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).

 

[FaraDazed]

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

Ah right, I see now! force times distance, I thought clever_name meant for me to double check what the question was asking lol.