# Recent content by Yae Miteo

1. ### Loop-the-loop with potentional and kinetic energy

Centripetal acceleration must be greater than gravitational acceleration.

KE \gt PE
3. ### Loop-the-loop with potentional and kinetic energy

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
4. ### Loop-the-loop with potentional and kinetic energy

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.
5. ### Loop-the-loop with potentional and kinetic energy

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...
6. ### Pendulum Velocity

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

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...
8. ### Spring launched box sliding over friction surface

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 =...
9. ### Work by tension on masses

Thank-you so much guys! I got it figured out.
10. ### Work by tension on masses

So then into W=Fd W=\cfrac{m_1m_2gd}{m_1 + m_2}
11. ### Work by tension on masses

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}
12. ### Work by tension on masses

And the 2nd law equations for tension are F_1=m_1a F_2=m_2g right?
13. ### Work by tension on masses

yes, it was a typo
14. ### Work by tension on masses

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?
15. ### Work by tension on masses

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