(a) The first sled stops momentarily when its kinetic energy is zero. At that time it has risen a vertical distance h such that
g h = Vo^2/2
h = Vo^2/(2g) = 2.13 m
distance up incline = h/sin39.8 = 3.33 m
(b) Compute the time it takes the first sled to go up (or down). Call that time T. Then compute how fast the second sled needs to be launched to get to the bottom in that time, traveling the full 15 m distance.
T*(Vo/2) = 3.33 m
T = 1.03 s
Take it from there
A frictionless plane is 15m long and inclined at 39.8 degrees. A sled starts at the bottom with an initial speed of 6.46m/s up the incline. When it reaches the point at which it momentarily stops, a second sled is released from the top of this incline with an initial speed vi. Both sleds reach the bottom of the incline at the same moment. The acceleration of gravity is 9.8m/s^2.
a) find the distance the the first sled traveled up the incline.
b) find the initial speed of the second sled
1 answer
1 answer