How to incorporate the initial vellocity

  • Thread starter Thread starter kbrowne29
  • Start date Start date
  • Tags Tags
    Initial
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the calculation of average force exerted by a shot-putter on a shot. The problem includes parameters such as mass, distance, and initial and final velocities, specifically focusing on how to incorporate initial velocity into the calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of kinematic equations to find acceleration and its relation to force. Questions arise regarding the definitions of variables and the implications of initial velocity on the calculations.

Discussion Status

Some participants have provided guidance on using kinematic equations, while others have expressed confusion regarding the application of acceleration in the force formula. There is an acknowledgment of mistakes in calculations, leading to a realization of the correct approach, although no consensus on the final answer has been reached.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for discussion. There is a specific focus on understanding the relationship between initial velocity, acceleration, and force.

kbrowne29
Messages
12
Reaction score
0
I'm having difficulty with the following problem: 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 2.8 m and is released with a speed of 13 m/s.

I know that I have to use the formula F=ma. The mass is obviously 7.0 kg, but I don't know how to find the horizontal component of the acceleration of the shot. I don't know how to incorporate the initial vellocity and the distance the shot travels. I would appreciate any help. Thanks.
 
Physics news on Phys.org
Use the formula:

v2 = u2 + 2as
 
Last edited:
what do the variables stand for?
 
v is final velocity (13ms-1), u is initial velocity (assume this to be zero), a is acceleration, s is displacement (distance).
 
ok, so that gives an acceleration of 2.32m/s^2. But it doesn't make sense that this would be the acceleration to plug into the formula F=ma. I know that the correct answer to the problem is 210 N.
 
Show me your workings, using the above equations I got an answer of 211.25 N (210 N if I'd rounded the acceleration to 3 ms-2).
 
Last edited:
I got it! I've been making such a dumb mistake. When I used v^2=u^2 + 2as, I kept forgetting to square the 13! Now I'm getting the right answer. Thanks a lot for helping me out here.
 

Similar threads

Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 157 ·
6
Replies
157
Views
14K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
5
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K