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? (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 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 mass is attached to a spring with a spring constant of 200 N/m. What is the force exerted by the spring when it is compressed by 0.1 m?
A.2 N
B.5 N
C.10 N
D.20 N
Solution
Using Hooke's Law, F = kx = 200 N/m * 0.1 m = 20 N.
Correct Answer: C — 10 N
Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum force exerted by the spring when it is compressed by 0.1 m?
A.2 N
B.5 N
C.20 N
D.10 N
Solution
Using Hooke's law, F = kx = 200 N/m * 0.1 m = 20 N.
Correct Answer: C — 20 N
Q. A 1 kg mass is attached to a spring with a spring constant of 200 N/m. What is the maximum extension of the spring when the mass is released from rest?
A.0.5 m
B.1 m
C.2 m
D.0.25 m
Solution
Using Hooke's law, F = kx, where F = mg = 1 kg * 9.8 m/s² = 9.8 N. Thus, x = F/k = 9.8 N / 200 N/m = 0.049 m.
Correct Answer: A — 0.5 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 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 1 kg object is pushed with a force of 10 N. If the frictional force is 4 N, what is the net force acting on the object?
A.6 N
B.10 N
C.4 N
D.0 N
Solution
Net force = applied force - frictional force = 10 N - 4 N = 6 N.
Correct Answer: A — 6 N
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 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 at rest on a horizontal surface. If a force of 30 N is applied, what is the object's acceleration assuming a frictional force of 10 N?
A.2 m/s²
B.3 m/s²
C.4 m/s²
D.5 m/s²
Solution
Net force = 30 N - 10 N = 20 N. Using F = ma, a = F/m = 20 N / 10 kg = 2 m/s².
Correct Answer: B — 3 m/s²
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 hanging from a rope. What is the tension in the rope when the object is at rest?
A.0 N
B.10 N
C.100 N
D.50 N
Solution
At rest, the tension in the rope equals the weight of the object: T = mg = 10 kg * 10 m/s² = 100 N.
Correct Answer: C — 100 N
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 lifted to a height of 5 m. What is the potential energy gained by the object?
A.50 J
B.100 J
C.150 J
D.200 J
Solution
Potential energy (PE) = mgh = 10 kg * 9.8 m/s² * 5 m = 490 J.
Correct Answer: B — 100 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 10 kg object is moving with a velocity of 3 m/s. What is its momentum?
A.10 kg·m/s
B.30 kg·m/s
C.3 kg·m/s
D.0.3 kg·m/s
Solution
Momentum p = mv = 10 kg * 3 m/s = 30 kg·m/s.
Correct Answer: B — 30 kg·m/s
Q. A 10 kg object is moving with a velocity of 5 m/s. What is its momentum?
A.10 kg·m/s
B.25 kg·m/s
C.50 kg·m/s
D.75 kg·m/s
Solution
Momentum p = mv = 10 kg * 5 m/s = 50 kg·m/s.
Correct Answer: C — 50 kg·m/s
Q. A 10 Ω resistor and a 20 Ω resistor are connected in series. What is the total resistance?
A.10 Ω
B.20 Ω
C.30 Ω
D.5 Ω
Solution
In series, the total resistance R = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.
Correct Answer: C — 30 Ω
Q. A 10-ohm resistor and a 20-ohm resistor are connected in series. What is the total resistance?
A.10 ohms
B.20 ohms
C.30 ohms
D.5 ohms
Solution
In series, the total resistance R = R1 + R2 = 10 + 20 = 30 ohms.
Correct Answer: C — 30 ohms
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 12 kg object is at rest on a horizontal surface. What is the normal force acting on it?
A.0 N
B.12 N
C.120 N
D.100 N
Solution
The normal force equals the weight of the object: N = mg = 12 kg * 10 m/s² = 120 N.
Correct Answer: C — 120 N
Q. A 15 kg cart is pushed with a force of 60 N. If the frictional force opposing the motion is 15 N, what is the acceleration of the cart?
A.2 m/s²
B.3 m/s²
C.4 m/s²
D.5 m/s²
Solution
Net force = 60 N - 15 N = 45 N. Using F = ma, a = F/m = 45 N / 15 kg = 3 m/s².
