Q. A ball is thrown vertically upwards with a speed of 30 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.45.9 m
B.46.0 m
C.46.1 m
D.46.2 m
Solution
Using conservation of energy, initial kinetic energy = potential energy at maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (30)²/(2 * 9.8) = 45.9 m.
Q. A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.14 m/s
B.20 m/s
C.10 m/s
D.5 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 10) = 14 m/s.
Q. A block of mass 2 kg is released from a height of 10 m. What is its speed just before it hits the ground?
A.0 m/s
B.10 m/s
C.14 m/s
D.20 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*10) = 14 m/s.
Q. A block of mass 2 kg is released from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.5 m/s
B.10 m/s
C.15 m/s
D.20 m/s
Solution
Using conservation of energy, potential energy at height = kinetic energy just before hitting the ground. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 5) = 10 m/s.
Q. A pendulum of length 2 m swings from a height of 1 m. What is the speed at the lowest point of the swing? (g = 9.8 m/s²)
A.4.4 m/s
B.3.1 m/s
C.2.8 m/s
D.5.0 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = sqrt(2gh) = sqrt(2 * 9.8 * 1) = 4.4 m/s.
Q. A pendulum swings from a height of 2 m. What is the speed at the lowest point of the swing?
A.2 m/s
B.4 m/s
C.6 m/s
D.8 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 2) = 4 m/s.
Q. A pendulum swings from a height of 5 m. What is the speed at the lowest point of the swing?
A.5 m/s
B.10 m/s
C.15 m/s
D.20 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*5) = 10 m/s.
Q. A roller coaster at the top of a hill has a potential energy of 5000 J. If it descends to a height of 10 m, what is its speed at the bottom? (g = 9.8 m/s²)
A.10 m/s
B.20 m/s
C.30 m/s
D.40 m/s
Solution
Using conservation of energy, initial PE + KE = final PE + KE. 5000 J = mgh + 0.5mv². Solving gives v = √(2(5000 - mgh)/m) = 30 m/s.
Q. A roller coaster starts from rest at a height of 30 m. What is its speed at the lowest point? (g = 9.8 m/s²)
A.10 m/s
B.15 m/s
C.20 m/s
D.25 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv². Solving gives v = √(2gh) = √(2 * 9.8 * 30) = 24.5 m/s.
Q. A roller coaster starts from rest at a height of 50 m. What is its speed at the lowest point?
A.10 m/s
B.20 m/s
C.30 m/s
D.40 m/s
Solution
Using conservation of energy, potential energy at the top = kinetic energy at the bottom. mgh = 0.5mv^2. Solving gives v = sqrt(2gh) = sqrt(2*9.8*50) = 31.3 m/s.
