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Lunna [17]
3 years ago
13

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

its kinetic energy
when it is halfway down the hill (15 m) at point B?
Physics
1 answer:
MrRissso [65]3 years ago
8 0

The kinetic energy halfway the hill is 2.86\cdot 10^5 J

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

U_A +K_A = U_B + K_B

where

U_A=mgh_A is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

g=9.8 m/s^2 (acceleration of gravity)

h_A = 30 m (height at point A)

K_A=\frac{1}{2}mv_A^2 is the initial kinetic energy, at point A , with

v_A=17 m/s (velocity at point A)

U_B=mgh_B is the final potential energy, at point B, where

h_B = 15 m (height at point B)

K_B=\frac{1}{2}mv_B^2 is the final kinetic energy, at point B, where

v_B is the velocity at point B

Here we are interested in finding K_B, so by re-arranging the equation and substituting we find:

K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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