What is the kinetic energy of the mass when mass reaches the bottom?

A 2 kg mass is moving with a speed of 1 m/s at the top of a frictionless slide that is 3 m high. What is the kinetic energy of the mass when it reaches the bottom? I believe this is the equation I'm supposed to use: KE=1/2mv2. (The second 2 is a squared). 2 things that really confuse me in this problem: the frictionless part and I don't know where to put the height. Thanks for the help!

Public Comments

1. This is a very good example why we use energies. At the top: kinetic energy Wk = 1/2 mv^2 = 1 potential energy Wp= mgh = 60 At the bottom all the potential energy gets transformed to kinetic (because none was lost to friction) Wk = 61 Wp = 0 What is beautiful with this aproach is that you do not care what kind of path the mass took in between : ) You could also very easily calculate the speed at the bottom. v = Sqrt(2 * Wk / m ) = Sqrt(61) = 7.* m/s
2. Protencel energy Gain = mgh = 2*9.8*3 since frictionless slide this is Converted to KE KE = 2*9.8*3= 58.8J
3. The rule to use is the conservation of energy. E(initial) = E(final) Mgh + 1/2Mv^2(initial) = 1/2Mv^2(final) (2kg*9.8m/s^2*3m) + .5*2kg*1m/s^2 = 59.8Joules
4. conservation of energy says that all energy is conserved so that means that the PE+KE on top of the table is equal to the KE at the bottom PE=mgh =(2)(9.8)(3) =58.8 J KE=.5mv^2 =.5(2)(1)^2 =(1)(1) =1 J PE+KE=58.8+1=59.8 J KE=.5mv^2 since KE has to equal 59.8 59.8=.5(2)v^2 59.8=(1)v^2 59.8/1=v^2 59.8=v^2 sqrt 59.8=v 7.73 m=v