A block with a mass of 15Kg is released from rest 10M above ground. From A to B it has 500J of Friction Force. A B it goes into a frictionless loop with a radius of 1.4 m. Energy @ B? Speed @ B, PE and KE at C?

Question

A block with a mass of 15Kg is released from rest 10M above ground.  From A to B it has 500J of Friction Force.  A B it goes into a frictionless loop with a radius of 1.4 m.  Energy @ B?  Speed @ B, PE and KE at C?

anonymous · Answer

Well, it's kinetic energy at B is just it's potential energy at A - 500J.

That is:
$$KE = mgh = 15kg*9.81\frac{m}{s^2}*10m$$

We can find it's speed at B by applying:
$$KE = \frac{1}{2}mv^2$$
and solve for v.

I need to know where C is to say anything about it.

anonymous · Answer

Oops.

$$KE = 15*9.81*10 - 500$$

shiraz14 · Answer

To sriggins5: Could you please illustrate your question with a diagram? For one thing, we are uncertain if A & B are on the same level horizontal surface or if there is a vertical displacement between them. Likewise, there is no mention of C in any part of the scenario described in the earlier part of the question until one reaches the last sentence (where C is mentioned 'out of the blue').

anonymous · Answer

|dw:1397788078623:dw|