⚡ Unit 3: Energy and Society — Practice Quiz

Assessment AS Learning — Self-Check (SPH3U)
🔄 Not Graded — Unlimited Retakes
Purpose: Test understanding of work, energy, conservation, power, efficiency, heat. Instant feedback with worked solutions.
Formulae: \( W=Fd\cos\theta \)  |  \( E_k=\tfrac{1}{2}mv^2 \)  |  \( E_g=mgh \)  |  \( P=W/t \)  |  \( \eta=E_\text{useful}/E_\text{input}\times 100\% \)  |  \( Q=mc\Delta T \)  |  \( c_\text{water}=4186 \) J/(kg·°C)  |  \( g=9.8 \) m/s²
Score: 0 / 10
Q1 — Work done by constant force
A worker pushes a \(50\)-kg crate \(8.0\) m horizontally with a constant horizontal force of \(120\) N. How much work does the worker do?
Answer: J
Solution:
\( W = Fd\cos(0°) = 120 \times 8.0 = 960 \) J
Q2 — Kinetic energy
Calculate the kinetic energy of a \(1200\)-kg car moving at \(25\) m/s.
Answer: J
Solution:
\( E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(1200)(25)^2 = \tfrac{1}{2}(1200)(625) = 3.75\times10^5 \) J
Q3 — Gravitational PE
A \(2.0\)-kg book is lifted from the floor to a shelf \(1.8\) m above. Calculate the change in gravitational potential energy.
Answer: J
Solution:
\( \Delta E_g = mgh = 2.0(9.8)(1.8) = 35.3 \) J
Q4 — Conservation of energy (drop)
A \(0.50\)-kg ball is dropped from rest at a height of \(20\) m. Ignoring air resistance, what is its speed just before hitting the ground?
Answer: m/s
Solution:
Conservation: \( mgh = \tfrac{1}{2}mv^2 \) → \( v = \sqrt{2gh} = \sqrt{2(9.8)(20)} = \sqrt{392} = 19.8 \) m/s
Q5 — Power
A motor lifts a \(40\)-kg load \(10\) m in \(8.0\) s at constant speed. Calculate the power output.
Answer: W
Solution:
\( W = mgh = 40(9.8)(10) = 3920 \) J. \( P = W/t = 3920/8 = 490 \) W
Q6 — Efficiency
A motor uses \(2000\) J of electrical energy and produces \(1500\) J of useful mechanical work. What is its efficiency?
Answer: %
Solution:
\( \eta = E_\text{useful}/E_\text{input} \times 100\% = 1500/2000 \times 100 = 75\% \)
Q7 — Heat / specific heat
How much heat is required to raise the temperature of \(0.50\) kg of water from \(20°\)C to \(80°\)C? (\( c=4186 \) J/(kg·°C))
Answer: J
Solution:
\( Q = mc\Delta T = 0.50(4186)(60) = 1.26\times 10^5 \) J
Q8 — Conservation with friction
A \(2.0\)-kg block slides from rest down a frictionless ramp of height \(5.0\) m, then slides on a rough horizontal surface (\( \mu_k = 0.20 \)) until it stops. How far does it travel on the horizontal surface?
Answer: m
Solution:
At bottom of ramp: \( E_k = mgh = 2.0(9.8)(5.0) = 98 \) J
Friction does work: \( W_f = f_k d = \mu_k mg d = 0.20(2.0)(9.8)d = 3.92d \)
Set \( 3.92d = 98 \) → \( d = 25 \) m
Q9 — STSE (concept)
Which Ontario electricity source has the highest typical efficiency at converting input energy to electrical output?
Solution:
Hydroelectric plants achieve about 85–90% efficiency because gravitational PE of water is converted directly to mechanical and then electrical energy with relatively few thermal losses.
Q10 — Calorimetry (heat exchange)
A \(0.20\)-kg block of metal at \(100°\)C is dropped into \(0.30\) kg of water at \(20°\)C. Final temperature is \(28°\)C. Calculate the specific heat capacity of the metal. (\( c_\text{water}=4186 \) J/(kg·°C))
Answer: J/(kg·°C)
Solution:
Heat lost by metal = heat gained by water:
\( m_M c_M (100-28) = m_W c_W (28-20) \)
\( 0.20 c_M (72) = 0.30(4186)(8) \)
\( 14.4 c_M = 10046 \) → \( c_M = 698 \) J/(kg·°C)