Q. A 2 kg ball is dropped from a height of 10 m. What is its speed just before it hits the ground? (g = 10 m/s²) (2023)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
15 m/s
Solution
Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 1/2 mv^2. Solving gives v = sqrt(2gh) = sqrt(2 * 10 m/s² * 10 m) = 14.14 m/s.
Q. A 2 kg block is sliding down a frictionless incline of height 5 m. What is its speed at the bottom?
A.
10 m/s
B.
5 m/s
C.
20 m/s
D.
15 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 * 5) = 10 m/s.
Q. A 2 kg block of metal at 100°C is placed in 1 kg of water at 20°C. What is the final temperature of the system? (Specific heat of water = 4.2 J/g°C, specific heat of metal = 0.9 J/g°C) (2021)
A.
30°C
B.
40°C
C.
50°C
D.
60°C
Solution
Using the heat transfer equation, the final temperature can be calculated to be 50°C.
Q. A 2 kg block slides down a frictionless incline of height 3 m. What is its speed at the bottom? (2023)
A.
6 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Solution
Using conservation of energy, Potential Energy at top = Kinetic Energy at bottom: mgh = 1/2 mv^2; v = sqrt(2gh) = sqrt(2 * 9.8 m/s² * 3 m) = 7.75 m/s (approx. 6 m/s)
Q. A 2 kg object is dropped from a height of 15 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
17.15 m/s
B.
12.25 m/s
C.
14.14 m/s
D.
10.0 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 * 15) = 17.15 m/s.
Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
14 m/s
B.
19.8 m/s
C.
20 m/s
D.
28 m/s
Solution
Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.
Q. A 2 kg object is dropped from a height of 20 m. What is its speed just before it hits the ground? (Assume g = 10 m/s²)
A.
10 m/s
B.
14 m/s
C.
20 m/s
D.
40 m/s
Solution
Using conservation of energy, Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5 mv². v = √(2gh) = √(2 × 10 m/s² × 20 m) = 20 m/s.
Q. A 2 kg object is dropped from a height of 5 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.
10 m/s
B.
7 m/s
C.
5 m/s
D.
15 m/s
Solution
Using conservation of energy: Potential Energy at height = Kinetic Energy just before hitting ground. mgh = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 5 m) = 10 m/s.
