Recent content by misyg

I set them equal to zero, the SPE, and came up with the equation for the top and middle of the hill...top(1/2mv2)+(mgh)=(1/2mv2)+(mgh)middle...(1.25v2)+(2.94)=(3.125)+(1.47) v2=1.324 vtop=1.1507 does that sound correct.

So W=0J-243,000J So it would take -243,000J of work to stop the car

Two carts, a blue one with a mass of 350g and a silver one with a mass of 500g, move down a frictionless, level track together at .20 m/s to the right. When the two carts reach the center of the track, the spring on the silver cart is released, and the two carts move apart. The final velocity of...

A 250g block slides down a frictionless hill. If the hill is 1.2m high and the speed of the block is 5 m/s when it is halfway down, what was the speed of the block at the bottom and at the top? I tried using the equation KE(top)+GPE(top)+SPE(top)=KE(middle)+GPE(middle)+SPE(middle) and the same...

While driving my 1,500kg car at 18 m/s, I unexpectedly see a stop sign and must brake hard. How much work must be done on the car to stop it? I used KE=1/2(mv2) I came up with the answer of 243,000 J, but my professor gave me only 5 out of 10 points for the answer...any one see where I went...