Recent content by Meco

Ok thanks again, hurray! One problem down!

Thanks for being so helpful! So when you do (0-3.833^2)/70 you get -2.0988=a ma=F 77*2.0988=161.6076=net force = μmg 161.6076= μ(754.6) 161.6076/754.6=μ μ=.21416 <--- looks a lot more reasonable :D What units is μ in?

So I tried to work it out like you said and came up with this: 77kg = mass 77*9.8=754.6 = the Normal force -.05125= the acceleration 77*-.05125 = net force = μmg so 77*-.05125 = -3.946= μ77(9.8) -3.946/754.6= μ -.0052279= μ

Ok so Vf=0 Vi=3.5876 Distance traveled=35 So (0-3.5876)/(70)= -.05125 F=77(-.05125) F=-3.94625N (Newtons right?) So now that I know force I put it into F(friction)=μk*N So -3.94625= μk* 77 <--- the Normal force is just the mass right? It is in Kg here, that feels wrong for some reason...

Spinnor I don't really understand what that equation is, maybe I was just taught with different variables. Angry, I am trying to follow what you are trying to tell me and am struggling. The mass of the two objects combined is 77kgs. The force of the combined system is what? The starting...

Ok, well the energy before and after equation is: (.03*5.5)+ (0.055*0)= (.03*V1')+(.055*V2') Momentum in X before: .030*5.5 Momentum in the Y before: 0 Momentum final for X must equal .165 Momentum final for Y must equal 0 I do not really understand how I can use this to find two...

Homework Statement A 65.0kg person, running horizontally with a velocity of +4.25 m/s, jumps onto a 12.0-kg sled that is initially at rest. 1) Ignoring the effects of friction during the collision, find the velocity of the sled and the person as they move away. 2) The sled and the person...

Homework Statement The drawing shows a collision between two balls. Ball A has a mass of .030kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with ball B, which has a mass of .055kg and is initially at rest. The collision is not head-on. After the collision...