anonymous
  • anonymous
a ball, starting from rest, requires a speed of 10 m/s when a force is applied for a distance of 40 m. if the ball has a mass of 5 kg, what is the force being applied?
Mathematics
chestercat
  • chestercat
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Okay, you can tell this is an energy problem, as opposed to an acceleration problem, because they give you force and distance as opposed to force and time.
anonymous
  • anonymous
So you want to first find the kinetic energy of the ball.\[\Large E_{kinetic}=\frac{1}{2}mv^2\]
anonymous
  • anonymous
Now work is just the change in energy. Since the ball started at rest, it started at 0 energy so\[\Large E_{work} = E_{kinetic} - 0 = E_{kinetic}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Work is force times distance so: \[\Large E_{work}=Fd\]\[\Large F = \frac{1}{d}E_{work}\]
anonymous
  • anonymous
Combine it all together and we get:\[F=\frac{1}{d}\frac{1}{2}mv^2\]Where \(d = 20m\), \(m=5kg\), and \(v=10m/s\)
anonymous
  • anonymous
@raytiller1 I couldn't have made it any easier for you without giving you the answer.
anonymous
  • anonymous
so \[\frac{ 1 }{ 20 }\frac{ 1 }{ 2 } 20(10)\]
anonymous
  • anonymous
@wio
anonymous
  • anonymous
You need to square velocity
anonymous
  • anonymous
You also need to look at my formula and understand what I did.

Looking for something else?

Not the answer you are looking for? Search for more explanations.