The problems are on the kinetic energy worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Kinetic energy is the energy an object has because it is moving, and it grows with both mass and speed. These worked answers correct the 20 problems of the grade 7 science worksheet on kinetic energy, from which store of energy? to high jump: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit in bold. Compare your method as well as your final number, because a correct method with a small slip still earns most of the credit in a real assessment.
Level: Grade 7 science (Year 8 · Class 7) · Curriculum: MS-PS3-1, MS-PS3-2
Problem 1 Which store of energy?
1. A rolling football is moving: kinetic energy.
2. A book on a high shelf is high above the ground: gravitational potential energy.
3. The diver is at rest but high up: gravitational potential energy.
4. A running dog is moving: kinetic energy.
Problem 2 Formulas and units
1. KE = ½ × m × v².
2. GPE = m × g × h (g = 9.8 N/kg).
3. Energy is measured in joules (J) and mass in kilograms (kg).
Problem 3 Heavy or fast?
1. The speeds are equal, so the larger mass wins: trolley Y has more kinetic energy (it has three times as much).
2. Speed doubles, so energy rises by 2² = four times (quadruples).
Problem 4 A kicked football
1. KE = ½ × m × v².
2. KE = ½ × 0.43 × 10² = 0.215 × 100 = 21.5 J.
Problem 5 Bag on a shelf
1. GPE = 2.0 × 9.8 × 1.5 = 29.4 J.
2. On the floor h = 0, so GPE = 0 J (taking the floor as the zero level).
Problem 6 Boxes on supports
1. P: 1.0 × 9.8 × 0.5 = 4.9 J. Q: 1.0 × 9.8 × 1.0 = 9.8 J. R: 1.0 × 9.8 × 2.0 = 19.6 J.
2. Box R has the most. 19.6 ÷ 4.9 = 4 times more than P, because its height is four times as large.
Problem 7 Mass in grams
1. 250 g ÷ 1000 = 0.25 kg.
2. GPE = 0.25 × 9.8 × 4.0 = 9.8 J.
Problem 8 Cyclist in km/h
1. 18 km/h = 18 000 m ÷ 3600 s = 5 m/s.
2. KE = ½ × 70 × 5² = 35 × 25 = 875 J.
Problem 9 Twice as fast
1. KE = ½ × 0.50 × 6² = 0.25 × 36 = 9 J.
2. KE = ½ × 0.50 × 12² = 0.25 × 144 = 36 J.
3. 36 ÷ 9 = 4 times larger. The speed is squared, so doubling the speed multiplies the energy by 2² = 4.
Problem 10 Reading an energy graph
1. At 3 m/s the energy is 9 J.
2. A kinetic energy of 16 J occurs at 4 m/s.
3. Kinetic energy depends on speed squared, so it increases faster and faster. A straight line would mean energy simply proportional to speed.
Problem 11 Height of a crate
1. h = GPE ÷ (m × g).
2. h = 147 ÷ (5.0 × 9.8) = 147 ÷ 49 = 3.0 m.
Problem 12 Falling apple
1. GPE lost = 0.50 × 9.8 × 2.0 = 9.8 J.
2. With no air resistance, all of it becomes kinetic energy: 9.8 J.
Problem 13 On a playground swing
1. At the highest point the swing is momentarily at rest, so it has the most gravitational potential energy and no kinetic energy.
2. At the lowest point the height is smallest and the speed is greatest: maximum kinetic energy, minimum potential energy.
3. Friction at the pivot and air resistance transfer some energy to the surroundings as thermal energy on every swing, so the swing rises less and less.
Problem 14 Scooter or car?
1. Scooter: ½ × 150 × 10² = 75 × 100 = 7500 J. Car: ½ × 1200 × 5² = 600 × 25 = 15 000 J.
2. The car has more, by 15 000 ÷ 7500 = 2 times.
Problem 15 Find the error
1. She left out the ½ in ½ × m × v², and doubling the speed does not double the energy because speed is squared.
2. At 3 m/s: ½ × 4 × 9 = 18 J. At 6 m/s: ½ × 4 × 36 = 72 J, which is four times 18 J.
Problem 16 Mass from energy
1. m = 2 × KE ÷ v².
2. m = 2 × 360 ÷ 6.0² = 720 ÷ 36 = 20 kg.
Problem 17 Speed of a cricket ball
1. v² = 2 × KE ÷ m.
2. v² = 2 × 30 ÷ 0.15 = 400, so v = √400 = 20 m/s.
Problem 18 Smooth track
1. GPE = 0.80 × 9.8 × 0.50 = 3.92 J.
2. The track is smooth, so all of it becomes kinetic: 3.92 J.
3. v² = 2 × 3.92 ÷ 0.80 = 9.8, so v = √9.8 = 3.1 m/s.
Problem 19 Stopping a car
1. At 20 m/s: ½ × 1000 × 400 = 200 000 J. At 30 m/s: ½ × 1000 × 900 = 450 000 J.
2. 450 000 ÷ 200 000 = 2.25 times.
3. The speed is only 1.5 times larger, but the energy is 1.5² = 2.25 times larger. The brakes must remove all of it, so the stopping distance grows much more than the speed.
Problem 20 High jump
1. KE = ½ × 60 × 7.0² = 30 × 49 = 1470 J.
2. h = GPE ÷ (m × g) = 1470 ÷ (60 × 9.8) = 1470 ÷ 588 = 2.5 m.
3. Not all kinetic energy can be converted: some is transferred as thermal energy in the muscles and the ground, and the jumper keeps some forward speed over the bar.
