ANSWERS: 2
  • First of all, what is the length of the ramp, its magnitude or its angle of inclination? Second of all, if the ramp is straight then the block CANNOT leave its end horizontally. If the ramp is curved then what is the equation of the curve?
  • The ramp could be like a part of a circle for instance. The block slides down a height 1.25m In doing so, it loses a potential energy of PE = m g h where m is mass, g is the gravity constant (about 10m/s/s) and h is 1.25m It gains the same amount of kinetic energy, KE = (1/2) m v^2 So (1/2) m v^2 = m g h v = sqrt(2 g h) It's not moving vertically when it leaves the ramp. It's vertical equation of motion is: s = (1/2) g t^2 where s is 0.25m The time taken to drop the rest of the way is t = sqrt(2s/g) if it is travelling at v horizontally, it will travel d= v t = v sqrt(2s/g) in that time. d = sqrt(2 g h) sqrt(2s/g) = 2 sqrt(h s)

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