Recent content by erm151

Ok I figured it out. I had to find out the velocity at the height of 0.8m then use the KE+PE=Total energy equation so that 1/2 mv^2+mgh=1/2mv^2+mgh and one side is the velocity at 0.8 and the other (the velocity I am looking for) is calculated at h=0. so 1/2(0.8132)(v^2)=1/2(0.8132)(v at...

So for the tension of the cord part would I have to include the cos(theta) into the work so T=mg(cos(theta)+ma? I don't understand what you mean by what I want to get out of the initial inelastic collision? If I understand you correctly I want to get the velocity of the bullet from the equality...

Homework Statement A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.60 m long. The block is initially at rest. A bullet with mass 0.0132 kg is fired at the block with a horizontal velocity v-_i. The bullet strikes the block and becomes embedded...