A roller coaster cart of mass m=351 kg starts stationary at point A, where h1= 29.2 m and a while later is at B, where h2= 10.1 m.
The acceleration of gravity is 9.8 m/s^2.
What is the speed of the cart at B, ignoring the effect of friction? Answer in units of m/s.
The vertical distance the roller coster cart of mass moved down is given by :
h1-h2= 29.2m-10.1m = 19.1m
There is no friction. We assume that the cart is moving only due to the gratational acceleration.
Let the acceleration due to gravity 9m/s^2 and the velocity of the cart V, the relation between is as follows:
v^2 = 2g(h1-h2)=2*9.81*19.1
v=sqrt(2*9.81*19.1)=19.3583m/s is the downward speed