Recent content by dch1runs

A particle of mass m moves from rest at t=0 under the influence of a single constant force F. Show that the power delivered by the force at any time t is P=(F^2t)/m. I tried using the definition of power and the definition of the scalar product, which probably is not the way to go about...

Should it be xcm= (m2*d)/(m2+m1) + .5?

1. You are standing at the very rear of a 6.0m long, 120kg raft that is at rest in a lake with its prow only .5m from the end of the pier. Your mass is 60kg. Neglect friction between the raft and the water. A) How far from the end of the pier is the center of mass of the you-raft system? B) You...

Alright, so I finally got it! Thank you, although I do have one general question with this type of problem. How is it that both objects have the same acceleration, is it because they are connected by the rope and pulley?

So for the 75g object the forces acting on it are gravity and the rope. F=m*a=m*g-T The 270g object also has gravity (though only the parallel component) and the rope. f=m*a=T-g*sin(theta) But I still can't find the answer when putting them together and solving for tension and acceleration...

Thank you for that. So now I can find the perpendicular and parallel components. The parallel component would be m*g*sin(theta). Which in this example is .27kg*9.81m/(s*s)*sin34.4 I think. But its still not the magnitude of the acceleration, right? I know the site said to divide by the mass...

Does that mean that it would be mg/cos34.4?

Its not given, but I think its 9.81cos34.4?

1. A frictionless surface is inclined at an angle of 34.4° to the horizontal. A 270-g block on the ramp is attached to a 75.0-g block using a pulley, as shown in the figure below. (a) Draw two free-body diagrams, one for the 270-g block and the other for the 75.0-g block. (b) Find the tension...

I can follow this problem through part (a), but I keep getting lost in part (b). Would someone mind clearing it up for me a little?