6. An object moving along a horizontal surface approaches the bottom of a ramp.

Question

6. An object moving along a horizontal surface approaches the bottom of a ramp. At the very bottom of the ramp, point A, its speed is 40 m/s. The object goes up the ramp and reverses direction at point B. The angle of inclination is 30°. Neglect friction. What is the height of point B above point A? (10 pts) A child throws a ball straight up, and gives the ball an initial speed of 8.00 m/s. The ball leaves her hand 1.00 m above the ground. What is the gravitational potential energy when the ball is at it's maximum height? (Make sure you define where gravitational potential energy is equal to zero) (10 pts) 7.

Explanation / Answer

total energy at point A = total energy at point B

KEA = PEB

(1/2)*m*v^A^2 = m*g*h

h = VA^2/(2g)

h = 40^2/(2*9.8)

h = 81.63 m

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7)

initial energy Ei = m*g*h + (1/2)*m*v^2

at the maximum height the speed = 0

the ball has only potential energy

total energy at maximum point Ef = PE

from energy conservation

Ef = Ei

PE = (1/2)*m*v^2 + m*g*h

PE = (1/2)*m*8^2 + m*9.8*1

PE = m*4.18 J

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