Work, Energy & Power
Q. A roller coaster at the top of a hill has a height of 30 m. If it has a mass of 500 kg, what is its potential energy? (g = 9.8 m/s²)
A.
14700 J
B.
9800 J
C.
15000 J
D.
5000 J
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Solution
Potential energy is given by PE = mgh. Here, PE = 500 * 9.8 * 30 = 14700 J.
Correct Answer: A — 14700 J
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Q. A roller coaster at the top of a hill has a height of 30 m. If it has a mass of 500 kg, what is its potential energy at the top?
A.
1500 J
B.
3000 J
C.
7500 J
D.
15000 J
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Solution
Potential energy = mgh = 500 * 9.8 * 30 = 147000 J.
Correct Answer: D — 15000 J
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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
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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.
Correct Answer: C — 30 m/s
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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
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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.
Correct Answer: C — 20 m/s
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Q. A roller coaster starts from rest at a height of 50 m. What is its speed at the lowest point? (g = 9.8 m/s²)
A.
10 m/s
B.
20 m/s
C.
30 m/s
D.
40 m/s
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Solution
Using conservation of energy, mgh = 0.5mv². Thus, v = sqrt(2gh) = sqrt(2 * 9.8 * 50) = 31.3 m/s, approximately 30 m/s.
Correct Answer: C — 30 m/s
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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
Show solution
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.
Correct Answer: C — 30 m/s
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Q. A spring with a spring constant of 200 N/m is compressed by 0.1 m. What is the potential energy stored in the spring?
A.
1 J
B.
2 J
C.
3 J
D.
4 J
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Solution
Potential Energy = 0.5 × k × x² = 0.5 × 200 N/m × (0.1 m)² = 1 J.
Correct Answer: A — 1 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.1 m. What is the work done in compressing the spring?
A.
1 J
B.
2 J
C.
3 J
D.
4 J
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Solution
Work done = 0.5 × k × x^2 = 0.5 × 200 N/m × (0.1 m)^2 = 1 J.
Correct Answer: A — 1 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.5 m. What is the potential energy stored in the spring?
A.
25 J
B.
50 J
C.
100 J
D.
200 J
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Solution
Potential Energy in spring = 0.5 × k × x² = 0.5 × 200 N/m × (0.5 m)² = 25 J.
Correct Answer: B — 50 J
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Q. A spring with a spring constant of 200 N/m is compressed by 0.5 m. What is the work done in compressing the spring?
A.
25 J
B.
50 J
C.
100 J
D.
200 J
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Solution
Work done = 0.5 × k × x² = 0.5 × 200 N/m × (0.5 m)² = 25 J.
Correct Answer: B — 50 J
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Q. An object is lifted to a height of 10 m. If the mass of the object is 5 kg, what is the gravitational potential energy gained?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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Solution
Potential Energy = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer: B — 100 J
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Q. An object is lifted vertically 10 m against gravity. If the mass of the object is 5 kg, what is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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Solution
Work done = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer: B — 100 J
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Q. An object is lifted vertically 3 m against gravity. If the mass of the object is 4 kg, what is the work done against gravity? (g = 9.8 m/s²)
A.
117.6 J
B.
39.2 J
C.
29.4 J
D.
19.6 J
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Solution
Work done = mgh = 4 kg × 9.8 m/s² × 3 m = 117.6 J.
Correct Answer: A — 117.6 J
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Q. An object is lifted vertically to a height of 10 m. If the mass of the object is 5 kg, what is the work done against gravity?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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Solution
Work done = mass × g × height = 5 kg × 9.8 m/s² × 10 m = 490 J.
Correct Answer: B — 100 J
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Q. An object is moving in a circular path with a constant speed. What can be said about the work done by the centripetal force?
A.
Positive work
B.
Negative work
C.
Zero work
D.
Depends on speed
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Solution
Centripetal force does no work as it acts perpendicular to the direction of motion.
Correct Answer: C — Zero work
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Q. An object is moving in a circular path with a constant speed. What type of energy is primarily involved?
A.
Kinetic Energy
B.
Potential Energy
C.
Mechanical Energy
D.
Thermal Energy
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Solution
The object has Kinetic Energy due to its motion in the circular path.
Correct Answer: A — Kinetic Energy
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Q. An object is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 10 m/s²)
A.
11.25 m
B.
22.5 m
C.
15 m
D.
20 m
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Solution
Maximum Height (h) = v^2 / (2g) = (15^2) / (2 * 10) = 11.25 m
Correct Answer: B — 22.5 m
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Q. If 10^(2x) = 100, what is the value of x?
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Solution
100 = 10^2, so 2x = 2, hence x = 1.
Correct Answer: B — 0.5
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Q. If 10^x = 0.01, what is the value of x?
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Solution
Since 0.01 = 10^(-2), we have x = -2.
Correct Answer: A — -2
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Q. If 2^(x+1) = 16, what is the value of x?
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Solution
16 = 2^4 => x + 1 = 4 => x = 3
Correct Answer: B — 3
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Q. If 3^(2x) = 81, what is the value of x?
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Solution
Since 81 = 3^4, we have 3^(2x) = 3^4, thus 2x = 4, giving x = 2.
Correct Answer: A — 2
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Q. If 4^x = 64, what is the value of x?
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Solution
Since 64 = 4^3, we have x = 3.
Correct Answer: B — 3
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Q. If 5^(x+1) = 125, what is the value of x?
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Solution
125 = 5^3, so x + 1 = 3, hence x = 2.
Correct Answer: B — 2
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Q. If 5^(x-1) = 25, what is the value of x?
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Solution
Since 25 = 5^2, we have 5^(x-1) = 5^2, thus x - 1 = 2, giving x = 3.
Correct Answer: A — 2
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Q. If 5^x = 125, what is the value of x?
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Solution
125 = 5^3 => x = 3
Correct Answer: C — 3
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Q. If 5^x = 25, what is the value of x?
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Solution
Since 25 = 5^2, we have 5^x = 5^2, thus x = 2.
Correct Answer: B — 2
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Q. If 7^(x) = 49, what is the value of x?
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Solution
49 = 7^2, so x = 2.
Correct Answer: B — 2
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Q. If 8^(x) = 64, what is the value of x?
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Solution
64 = 8^2, so x = 2.
Correct Answer: B — 2
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Q. If a 10 kg object is dropped from a height of 20 m, what is its potential energy at the top? (1960)
A.
200 J
B.
1000 J
C.
500 J
D.
1500 J
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Solution
Potential Energy (PE) = m * g * h = 10 kg * 9.8 m/s^2 * 20 m = 1960 J
Correct Answer: B — 1000 J
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Q. If a 10 kg object is dropped from a height of 5 m, what is the potential energy at that height?
A.
50 J
B.
100 J
C.
150 J
D.
200 J
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Solution
Potential energy = mgh = 10 kg × 9.8 m/s² × 5 m = 490 J.
Correct Answer: B — 100 J
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