A 200 g particle gains a momentum of 25 Ns in 12 s starting from rest. Find the kinetic energy of the particle and the applied force.

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 our expert's

answer on brainly

SEE EXPERT ANSWER

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

A community for students.

A 200 g particle gains a momentum of 25 Ns in 12 s starting from rest. Find the kinetic energy of the particle and the applied force.

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

We are given the mass of the particle m=200 g = 0.2 kg. And the momentum p = 25 Ns = 25 kg m/s. We know \(p=mv\) and kinetic energy \(E_k = \frac{1}{2}mv^2\). Multiplying numerator and denominator of the expression for \(E_k\) by m, we get: \[E_k = \frac{p^2}{2m} \] Plugging in our given quantities, yields: \[E_k = 1562.5\ \mathrm{J}\] Assuming a constant force, we can use the impulse/momentum equation: \[F\Delta t = \Delta p\] Since we are starting from rest \(\Delta p = p\), and we are given \(\Delta t = 12\mathrm{s}\). So that: \[F = \frac{p}{\Delta t} = \frac{25\ \mathrm{Ns}}{12s} = 2.08\ \mathrm{N}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question