Recent content by Joe Armas

So Friction force on C is pointing up the ramp and its magnitude is equal to friction force on B? The frictional force on C is kinetic not static correct?

Homework Statement In the picture Homework Equations Torque = I * Alpha Alpha = a / r Fnet = m*a The Attempt at a Solution I provided my work in the pictures. I figured out that the frictional force in system B is greater than in system A. I am not sure how to compare this with the...

The system consists of the two gliders. So it appears that the spring is indeed an external force. But the direction the spring accelerates the block is in negative direction, so I am not sure how the TOTAL momentum of the system will change.

Homework Statement In the pictures Homework Equations Fnet = dp/dt The Attempt at a Solution I know in part b there is a net external force (gravity) that does positive work, fnet > 0, thus dp/dt is positive. Does the spring in part c increase or decrease the momentum of the system? I am...

I know in part b there is a net external force (gravity) that does positive work, fnet > 0, thus dp/dt is positive. Does the spring in part c increase or decrease the momentum of the system? I am not too sure of my answer.

Is this reasoning correct?

Gravity then should be negative since it too acts downward. So, delta p should be -30 * 0.4 or -12? So final momentum of system is actually -24 kg * m/s. The bottom piece of mass 2kg must therefore be moving at -14 m/s, since the top piece of mass 1kg is moving at +4m/s, and the sum of these...

The initial momentum of the system is 3kg * -4m/s = -12 kg * m/s. The change in momentum from gravity is 12 as I explained. Pf = Pi + Delta P Pf = -12 + 12 = 0

Homework Statement In the picture Homework Equations In the picture attached. The Attempt at a Solution For the firecracker problem, I believe the only net external force is gravity. Therefore using, Fnet * delta t = delta p. Change in momentum is 12 kg * m/s. Thus, the final momentum of the...

The system gains potential energy because the spring is stretched. Isn't Wnet external equal to - delta PE? So it would be indeed be negative? That was my reasoning and it makes sense.

What situations can we use conservation of energy for net external work?

It should be positive right? The string does positive work while work does negative work. But why is the work net external not equal to delta potential energy here?

Ok thanks for the help. Do you see any discrepancies on the back side?

The wall does zero work while the String does positive work right? Since displacement and force is in the same direction?

No, correct? The only force that I can see is the wall.