Recent content by Yae Miteo

Centripetal acceleration must be greater than gravitational acceleration.

KE \gt PE

Hmm... if it had zero velocity at the top, it would stop and fall straight down. I need to develop a different approach. Perhaps at the top

It needs enough kinetic energy to reach the top, which gets turned into potential energy right when it gets there. That way, it will make it around the loop.

Homework Statement A mass m = 77 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.1 m and finally a flat straight section at the same height as the center of the loop (15.1 m off the ground). Since the mass would not make it around the loop if...

The initial energy will all be potential (PE = mgh) and the final energy will be entirely kintic (KE = 1/2 mv^2)

Homework Statement A mass m = 5.5 kg hangs on the end of a massless rope L = 1.81 m long. The pendulum is held horizontal and released from rest. How fast is the mass moving at the bottom of its path? Homework Equations a_c = \frac {v^2}{r} F = ma v = v_o + at The Attempt at a Solution I...

Homework Statement A block with mass m = 14 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4085 N/m after being compressed a distance x_1 = 0.546 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d =...

Thank-you so much guys! I got it figured out.

So then into W=Fd W=\cfrac{m_1m_2gd}{m_1 + m_2}

I worked it like this: tension will equal the sum of both forces, so m_2a + m_2g = T and T = m_1a solve for a a = \cfrac{m_2g}{m_1 + m_2} plug in T = \cfrac{m_1m_2g}{m_1+m_2}

And the 2nd law equations for tension are F_1=m_1a F_2=m_2g right?

yes, it was a typo

So, solving for tension, I get T = \cfrac{m_1m_2g}{m_1-m_2} and then putting it into W=Fd I get W=\cfrac{m_2m_2gd}{m_1+m_2} Is this correct?

Would tension then be T=m_1g or would it involve both mass 1 and mass 2?