Question

1. If a roller coaster has a total of 6396 Joules of energy at the top of a hill, then at the bottom of the hill how much total energy will it have.

(assume: no brakes, no motor, no air resistance, no external forces)

Hint: use conservation of energy!

Question 2

If a roller coaster begins at rest at the top of a 20 meter hill, then at the bottom of the hill how fast will it be going?

Hint: PE at the top = KE at the bottom

If a roller coaster begins at rest at the top of a hill, and at the bottom of the hill is moving at 38 m/s, how tall was the hill?

Hint: PE at the top = KE at the bottom

Question 4

A roller coaster begins at rest at the top of a 25 m hill. Find the velocity at a point 6 m above the bottom of the hill.

Hint: PE at point 1 = (KE + PE) at point 2

Question 3

If a roller coaster begins at rest at the top of a hill, and at the bottom of the hill is moving at 38 m/s, how tall was the hill?

Question 5

A roller coaster begins at 7 m/s at the top of a 38 m hill. Find the velocity at a point 10 m above the bottom of the hill.

Hint: (KE + PE) at point 1 = (KE + PE) at point 2

Question 6

A book is dropped from a height of 25 m. Find its speed when it is halfway down.

7. An arrow is shot straight up into the sky at a velocity of 45 m/s. Use conservation of energy to find the maximum height.

8. How much force does a car engine exert to change the speed of a 977 kg car from 8 m/s to 30 m/s in s distance of 306 m

HINT: notice force, distance, change of velocity

W = ΔKE

Fx = 1/2 mv₂² - 1/2 mv₁²

Question 9

Find the distance it will take a car engine to change the speed of a 1338 kg car from 10 m/s to 28 m/s if it exerts a force of 2245 N

HINT: notice force, distance, change of velocity

W = ΔKE

Fx = 1/2 mv₂² - 1/2 mv₁²

Answer 1

1) The total energy is conserved. At the bottom of the hill, the total energy will be 6396 J

2) potential energy at the top of the hill = kinetic energy at the bottom of the hill

mgh = mv^2/2

or speed at the bottom of the hill is

v = sqrt(2gh) = sqrt(2 x 9.8 x 20) = 19.8 m/s

3) using mgh = mv^2/2

or the height of the hill is

h = v^2/(2g) = 38^2/(2x9.8) = 73.7 m

4) 6 m above the bottom is h = 25-6 = 19 m

Therefore speed at 6 m above the bottom of the hill is

v= sqrt(2 x 9.8 x 19) = 19.3 m/s

5) initial velocity u = 7 m/s

mu^2/2 + m gh= + mv^2/2

distance covered by the roller coaster 10m above the bottom is h = 38-10 = 28 m

or the final velocity at 10 m above the bottom of the hill is

v = sqrt(u^2+2gh) = sqrt(7^2+2 x 9.8 x 28) = 24.45 m/s

6) halfway down is h = 25/2 = 12.5 m

velocity halfway down is

v = sqrt(2 gh) = sqrt(2 x 12.5 x 9.8) = 15.65 m/s

7) maximum height of the arrow is

h = v^2/2g = 45^2/(2 x 9.8) = 103.3 m

8) change in energy is

F x = mv^2/2 -mu^2/2 = 977(30^2-8^2)/2 = 408386 J

or force exerted by the car engine is

F = 408386/x = 408386/306 = 1334.6 N

9) change in energy is

F x = mv^2/2 -mu^2/2 = 1338(28^2-10^2)/2 = 457596 J

distance covered by the car is

x = 457596/F = 457596/2245 = 203.8 m