10 questions · Form 3 Science Bab 7: Energy and Power
10 questions · plays in fullscreen
Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.
1. A rubber bowstring is pulled back with an average force of $100 N$ by a distance of $0.4 m$. How much elastic potential energy is stored in the bowstring?
Answer: A
$E_e = 12 F x = 12 × 100 N × 0.4 m = 20 J$.
2. A student weighing $450 N$ climbs a staircase of vertical height $4 m$ in $6 seconds$. What is the useful power generated by the student?
Answer: B
Work $W = F × s = 450 N × 4 m = 1800 J$. Power $P = 18006 = 300 W$.
3. What happens to the potential energy of a roller coaster cart as it descends down a steep track slope?
Answer: B
As height decreases, gravitational potential energy decreases while converting directly into kinetic energy (speeding up the coaster).
4. An electric motor takes 4 seconds to perform 800 J of work. What is the power output of the motor?
Answer: D
Power $P = Wt = 800 J4 s = 200 W$.
5. What is the rate at which work is done or energy is transferred called?
Answer: B
Power is defined as the rate of doing work ($P = Wt$).
6. A worker pushes a crate with a constant horizontal force of $50 N$ across a smooth floor over a distance of $4 m$. How much work is done by the worker?
Answer: C
Work $W = F × s = 50 N × 4 m = 200 J$.
7. What is the formula used to calculate the kinetic energy ($E_k$) of a moving object?
Answer: C
Kinetic energy of a mass $m$ moving at speed $v$ is given by $E_k = 12 m v2$.
8. A $1 kg$ rock is dropped from rest from a height of $20 m$. What is its kinetic energy right before it hits the ground? (Ignore air resistance, $g = 10 m s-2$)
Answer: C
Gain in Kinetic Energy = Loss in Gravitational Potential Energy = $mgh = 1 × 10 × 20 = 200 J$.
9. In which of the following scenarios is NO work done according to physics definition?
Answer: B
Work requires displacement in the direction of applied force ($W = F × s$). Since displacement is zero ($s = 0$), no work is done.
10. Using conservation of energy, calculate the velocity of a $0.5 kg$ ball dropped from a height of $5 m$ just before hitting the floor ($g = 10 m s-2$).
Answer: A
$mgh = 12 m v2 arrow v = √2gh = √2 × 10 × 5 = √100 = 10 m s-1$.