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Finding Work Done, Acceleration, Force and Time

  • Thread starter FaraDazed
  • Start date
  • #1
347
2

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:
[tex]
W=\frac{1}{2}m(v^2 - u^2) \\
W=\frac{1}{2}2(25^2 - 15^2) \\
W= 25^2 - 15^2 = 400J
[/tex]

Part B:
[tex]
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
[/tex]

Part C
[tex]
v=u+at \\
∴ t = \frac{v-u}{a} \\
t= \frac{25-15}{25} = 0.4s
[/tex]

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 :).
 

Answers and Replies

  • #2
125
0
look at the definition of work, for part b.
 
  • #3
265
0
For part (c), you can use the impulse-momentum theorem to check your answer
 
  • #4
347
2
look at the definition of work, for part b.
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
347
2
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
[tex]
I=Ft \\
I=50 \times 0.4 = 20Ns \\

Δp=mv-mu \\
Δp=2 \times 25 - 2 \times 15 = 50-30 = 20Ns
[/tex]

Does that mean it is correct?
 
  • #6
265
0
Are right, is that that the impulse is equal to the change in momentum?

that would be then
[tex]
I=Ft \\
I=50 \times 0.4 = 20Ns \\

Δp=mv-mu \\
Δp=2 \times 25 - 2 \times 15 = 50-30 = 20Ns
[/tex]

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).
 
  • #7
347
2
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.
 

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