Machines B/C

[Umaroth]

A ball with mass 7.50 kg reaches the base of a ramp with a slope of 30.0 degrees at a velocity of 15.0 m/s. If it rolls up a distance of 5.75 m before stopping, what is the coefficient of friction of the ramp?

[AlfWeg]

A ball with mass 7.50 kg reaches the base of a ramp with a slope of 30.0 degrees at a velocity of 15.0 m/s. If it rolls up a distance of 5.75 m before stopping, what is the coefficient of friction of the ramp?

About 1.73? By sum of forces or energy...

[Umaroth]

A ball with mass 7.50 kg reaches the base of a ramp with a slope of 30.0 degrees at a velocity of 15.0 m/s. If it rolls up a distance of 5.75 m before stopping, what is the coefficient of friction of the ramp?

About 1.73? By sum of forces or energy...

Ah, I messed up, I meant 15.75 m, not 5.75, you did it correctly but I messed up the problem. With 15.75 m you should get 0.264.

[shrewdPanther46]

EDIT nvm

[AlfWeg]

A ramp is 5 meters high and 13 meters long. A solid sphere is released at the top of the inclined plane. Friction is negligible. Calculate velocity at the bottom of the ramp. Assume g = 10
m/s^2. Edit: I made a mistake in this question

[Umaroth]

A ramp is 5 meters high and 13 meters long. A solid sphere is released at the top of the inclined plane. Friction is negligible. Calculate velocity at the bottom of the ramp. Assume g = 10
m/s^2. ans is not V=sqrt(2gh)

10 m/s

[AlfWeg]

A ramp is 5 meters high and 13 meters long. A solid sphere is released at the top of the inclined plane. Friction is negligible. Calculate velocity at the bottom of the ramp. Assume g = 10
m/s^2. ans is not V=sqrt(2gh)

10 m/s

Edit: Made a mistake in the question. Ur right!
Your turn!

[shrewdPanther46]

A ramp is 5 meters high and 13 meters long. A solid sphere is released at the top of the inclined plane. Friction is negligible. Calculate velocity at the bottom of the ramp. Assume g = 10
m/s^2. ans is not V=sqrt(2gh)

10 m/s

I think you missed that it's a sphere and it's rolling, so we have to include rotational kinetic energy in our energy equation. Taking that into Consideration, We get [KE (translational) + KE (Rotational) = PE] The formula for rotational KE is (1/2 * Rotational Inertia * Omega^2 ). A solid sphere has a rpt. Inertia of (2/5) * Mass * (Radius^2) and Omega = V/R. Thus, we get the equation
(1/2) * M * V^2 + (1/2) * (2/5) *M * R^2 = M* G*H. Rearranging we get, v = sqrt((10/7)gh), giving us 8.5 m/s^2. This Presentation might help to understand this. Your turn!

Umaroth's solution is correct, the sphere has no reason to roll

[AlfWeg]

10 m/s

I think you missed that it's a sphere and it's rolling, so we have to include rotational kinetic energy in our energy equation. Taking that into Consideration, We get [KE (translational) + KE (Rotational) = PE] The formula for rotational KE is (1/2 * Rotational Inertia * Omega^2 ). A solid sphere has a rpt. Inertia of (2/5) * Mass * (Radius^2) and Omega = V/R. Thus, we get the equation
(1/2) * M * V^2 + (1/2) * (2/5) *M * R^2 = M* G*H. Rearranging we get, v = sqrt((10/7)gh), giving us 8.5 m/s^2. This Presentation might help to understand this. Your turn!

Umaroth's solution is correct, the sphere has no reason to roll

Oh lol I forgot to say it was rolling. Ignore my stupidity

[terence.tan]

what is the ideal mechanical advantage of an Archimedes screw?