A roller coaster with a mass of 500.0 kg starts from a height of 27.0 meters with a speed of 5.1 m/s. It descends to a height of 2.00 meters before it starts back up. Some energy is loss to the work due to friction. The speed at the bottom of the ramp is 22.0 m/s. How much energy was lost to friction?
4 answers
KEtop + (m g h) = KEbottom + (friction loss)
I'm not getting any correct answers with this formula....I'm not sure what I'm doing wrong here
1/2(0.5)(5.1)^2 top
1/2(0.5)(22.0)^2 bottom
what?
1/2(0.5)(5.1)^2 top
1/2(0.5)(22.0)^2 bottom
what?
how do I calculate friction?
KE top + PE top -energy lost to friction (L)=KE bottom + PE bottom
500*9.8*27 + 1/2(500)*5.1^2 =138802.5
500*9.8*2 + 1/2(500)*22^2 = 130800
138802.5-L=130800
138802.5=130800+ L
138802.5-130800=L
8002 J lost to friction
500*9.8*27 + 1/2(500)*5.1^2 =138802.5
500*9.8*2 + 1/2(500)*22^2 = 130800
138802.5-L=130800
138802.5=130800+ L
138802.5-130800=L
8002 J lost to friction
1 answer