# Calculate F given m, d, and v should be easy.

1. Jun 12, 2012

### TrpnBils

1. The problem statement, all variables and given/known data
What is the average force exerted by a shot putter on a 7.0-kg shot if the shot is moved through a distance of 2.8m and is released with a speed of 13m/s?

2. Relevant equations
F = m * a
a = Δv/t
v = Δd/t

3. The attempt at a solution

This is a problem that should be easy, I think. I get one answer, but the answer in the back of the book for this particular problem is way different.

t = d/v = 2.8m/13m/s = 0.22 seconds

a = v/t = 13m/s / 0.22s = 59.09 m/s/s

F = ma = 7kg * 59.09 m/s/s = 413.6 N....right?

My calculation of 413.6N doesn't agree with the book's 210N in the answer key....

2. Jun 12, 2012

### azizlwl

You have to use average velocity not final velocity.
Or just constant acceleration(average force) with final velocity, 13m/s

Last edited: Jun 12, 2012
3. Jun 12, 2012

### Muphrid

Are you familiar with work and kinetic energy? If so, setting the kinetic energy $mv^2/2$ equal to the work done by the force $Fd$ would be an appropriate method.

4. Jun 12, 2012

### TrpnBils

That makes sense.... so my answer given above would be the instantaneous force right at the release rather than the average throughout the whole thing?

5. Jun 12, 2012

### azizlwl

I guess it's not.
What we are given the distance travel from 0 to final velocity.
If we apply constant force(acceleration), we will find the time taken to cover that distance.
The actual force applied is not linear.

Average velocity x time=distance

Last edited: Jun 12, 2012
6. Jun 14, 2012

### cool_jessica

final kin energy is E=0.5m*v^2, where m=7kg, v=13m/s;
work done on the shot is E=F*s, where s=2.8m, F is force by putter;
energy conserves; E=E; 0.5m*v^2= F*s, hence
F=0.5m*v^2 / s =

7. Jun 14, 2012

### Curious3141

Mathematically, the average force is given by:

$$\bar{F} = \frac{1}{s}\int_0^s Fdx$$

The definite integral is work done on the object, which is equal to the final kinetic energy, which is why you end up with:

$$\bar{F} = (\frac{1}{s})(\frac{1}{2}mv^2)$$.

8. Jun 14, 2012

### gbaby370

Simplest way to solve this problem,

Use work energy theorem,

fcosθd=(0.5)mv^2 (cosθ= 1 assuming all force is in the same direction as the displacement)

Rearrange algebraicly and you'll get the answer.

Hope this helps!

9. Jun 14, 2012

### dimension10

I thought the same!

10. Jun 14, 2012

### azizlwl

http://img864.imageshack.us/img864/4082/collision.jpg [Broken]

http://img256.imageshack.us/img256/5251/lolljg.jpg [Broken]

Last edited by a moderator: May 6, 2017