Q. A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches?
A.5 m
B.10 m
C.15 m
D.20 m
Solution
Using energy conservation, mgh = 0.5 mv², h = v²/(2g) = (10 m/s)²/(2 × 9.8 m/s²) = 5.1 m.
Correct Answer: A — 5 m
Q. A 1 kg ball is thrown vertically upwards with a speed of 10 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.5 m
B.10 m
C.15 m
D.20 m
Solution
Using energy conservation, initial kinetic energy = mgh. 0.5 × 1 kg × (10 m/s)² = 1 kg × 10 m/s² × h. h = 5 m.
Correct Answer: B — 10 m
Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.20.4 m
B.30.4 m
C.40.8 m
D.50.0 m
Solution
Using energy conservation, initial kinetic energy = mgh; 0.5 × 1 kg × (20 m/s)² = 9.8 m × h; h = 20.4 m.
Correct Answer: A — 20.4 m
Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?
A.10 m
B.20 m
C.30 m
D.40 m
Solution
Using energy conservation: KE_initial = PE_max; 0.5 × m × v² = mgh; h = v²/(2g) = (20 m/s)²/(2 × 9.8 m/s²) = 20.4 m.
Correct Answer: B — 20 m
Q. A 1 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.20 m
B.30 m
C.40 m
D.50 m
Solution
Using energy conservation: Initial K.E. = Potential Energy at max height. 0.5mv² = mgh. h = v²/(2g) = (20 m/s)² / (2 × 10 m/s²) = 20 m.
Correct Answer: B — 30 m
Q. A 1 kg mass is dropped from a height of 1 m. What is its speed just before it hits the ground?
A.1 m/s
B.2 m/s
C.3 m/s
D.4 m/s
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 1) = 4.43 m/s.
Correct Answer: B — 2 m/s
Q. A 1 kg mass is dropped from a height of 10 m. What is its speed just before it hits the ground?
A.5 m/s
B.10 m/s
C.14 m/s
D.20 m/s
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 10) = 14 m/s.
Correct Answer: C — 14 m/s
Q. A 1 kg mass is lifted to a height of 5 m. How much work is done against gravity?
A.5 J
B.10 J
C.15 J
D.20 J
Solution
Work done against gravity = mgh = 1 * 9.8 * 5 = 49 J.
Correct Answer: B — 10 J
Q. A 10 kg object falls freely from a height of 20 m. What is its potential energy at the top?
A.100 J
B.200 J
C.300 J
D.400 J
Solution
Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.
Correct Answer: B — 200 J
Q. A 10 kg object falls freely from a height of 20 m. What is its speed just before hitting the ground? (g = 9.8 m/s²)
A.10 m/s
B.14 m/s
C.20 m/s
D.28 m/s
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 28 m/s.
Correct Answer: D — 28 m/s
Q. A 10 kg object falls from a height of 20 m. What is its potential energy at the top?
A.100 J
B.200 J
C.300 J
D.400 J
Solution
Potential energy is given by PE = mgh = 10 * 9.8 * 20 = 1960 J.
Correct Answer: D — 400 J
Q. A 10 kg object falls from a height of 20 m. What is its speed just before it hits the ground? (g = 9.8 m/s²)
A.10 m/s
B.14 m/s
C.20 m/s
D.28 m/s
Solution
Using conservation of energy, v = √(2gh) = √(2 * 9.8 * 20) = 19.8 m/s.
Correct Answer: B — 14 m/s
Q. A 10 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 = 0.5mv². v = √(2gh) = √(2 × 9.8 m/s² × 20 m) = 19.8 m/s.
Correct Answer: B — 19.8 m/s
Q. A 10 kg object is lifted to a height of 2 m. What is the work done against gravity?
A.20 J
B.40 J
C.60 J
D.80 J
Solution
Work done = mass × g × height = 10 kg × 9.8 m/s² × 2 m = 196 J.
Correct Answer: B — 40 J
Q. A 10 kg object is lifted to a height of 5 m. How much work is done against gravity?
A.50 J
B.100 J
C.150 J
D.200 J
Solution
Work done = mgh = 10 * 9.8 * 5 = 490 J.
Correct Answer: B — 100 J
Q. A 10 kg object is moving with a speed of 5 m/s. What is its total mechanical energy?
A.125 J
B.250 J
C.500 J
D.1000 J
Solution
Total mechanical energy = KE + PE. KE = 0.5 * 10 * (5)² = 125 J. Assuming PE = 0, total energy = 125 J.
Correct Answer: B — 250 J
Q. A 10 kg object is moving with a velocity of 3 m/s. What is its kinetic energy?
A.15 J
B.30 J
C.45 J
D.60 J
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 10 kg × (3 m/s)² = 45 J.
Correct Answer: C — 45 J
Q. A 1000 kg car accelerates from rest to a speed of 20 m/s in 10 seconds. What is the average power exerted by the engine?
A.2000 W
B.4000 W
C.5000 W
D.6000 W
Solution
First, calculate the work done: W = (1/2)mv^2 = (1/2)(1000 kg)(20 m/s)^2 = 200000 J. Then, power is P = W/t = 200000 J / 10 s = 20000 W.
Correct Answer: C — 5000 W
Q. A 1000 W heater operates for 1 hour. How much energy does it consume?
A.3600 J
B.1000 J
C.3600000 J
D.100000 J
Solution
Energy consumed is E = P * t. Here, P = 1000 W and t = 1 hour = 3600 seconds. Thus, E = 1000 W * 3600 s = 3600000 J.
Correct Answer: C — 3600000 J
Q. A 1000 W heater operates for 3 hours. How much energy does it consume?
A.3000 J
B.1080000 J
C.3600000 J
D.1000 J
Solution
Energy consumed is E = P * t. Here, P = 1000 W and t = 3 hours = 10800 seconds. Thus, E = 1000 W * 10800 s = 10800000 J.
Correct Answer: C — 3600000 J
Q. A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches?
A.10 m
B.20 m
C.30 m
D.40 m
Solution
Using conservation of energy, KE at launch = PE at max height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2*9.8) = 20.41 m.
Correct Answer: B — 20 m
Q. A 2 kg ball is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 9.8 m/s²)
A.20.4 m
B.30.4 m
C.40.4 m
D.50.4 m
Solution
Using conservation of energy, KE at the bottom = PE at the maximum height. 0.5mv² = mgh. Solving gives h = v²/(2g) = (20)²/(2 * 9.8) = 20.4 m.
Correct Answer: B — 30.4 m
Q. A 2 kg object is dropped from a height of 10 m. What is its potential energy at the top?
A.20 J
B.40 J
C.60 J
D.80 J
Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 10 m = 196 J.
Correct Answer: B — 40 J
Q. A 2 kg object is dropped from a height of 10 m. What is the speed of the object just before it hits the ground? (g = 9.8 m/s²)
A.10 m/s
B.14 m/s
C.20 m/s
D.30 m/s
Solution
Using energy conservation, mgh = 0.5mv²; v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 10 m) = 14 m/s.
Correct Answer: B — 14 m/s
Q. A 2 kg object is dropped from a height of 20 m. What is its potential energy at the top? (g = 9.8 m/s²)
A.39.2 J
B.196 J
C.78.4 J
D.98.0 J
Solution
Potential Energy (PE) = m × g × h = 2 kg × 9.8 m/s² × 20 m = 392 J.
Correct Answer: B — 196 J
Q. A 2 kg object is dropped from a height of 5 m. What is its potential energy at the top?
A.10 J
B.20 J
C.30 J
D.40 J
Solution
Potential Energy = mass × g × height = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the height?
A.10 J
B.20 J
C.30 J
D.40 J
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 20 J
Q. A 2 kg object is dropped from a height of 5 m. What is the potential energy at the top? (g = 9.8 m/s²)
A.19.6 J
B.39.2 J
C.49 J
D.98 J
Solution
Potential energy = mgh = 2 kg × 9.8 m/s² × 5 m = 98 J.
Correct Answer: B — 39.2 J
Q. A 2 kg object is moving with a speed of 3 m/s. What is its kinetic energy?
A.6 J
B.9 J
C.12 J
D.18 J
Solution
Kinetic energy = 0.5 × m × v² = 0.5 × 2 kg × (3 m/s)² = 9 J.
Correct Answer: B — 9 J
Q. A 2 kg object is moving with a speed of 3 m/s. What is its total mechanical energy?
A.9 J
B.12 J
C.15 J
D.18 J
Solution
Total mechanical energy = kinetic energy + potential energy. KE = 0.5mv² = 0.5 * 2 * (3)² = 9 J.
