⚙️ Unit 2: Forces — Practice Quiz

Assessment AS Learning — Self-Check (SPH3U)

🔄 Not Graded — Unlimited Retakes

Purpose: Test understanding of Newton's laws, FBDs, friction, inclined planes, and connected bodies. Instant feedback with worked solutions.

Score: 0 / 10

Q1 — Net force

A \(5.0\)-kg object experiences forces of \(20\) N east and \(8\) N west. What is the magnitude and direction of acceleration?

Answer: m/s² east

Solution:

\( F_\text{net} = 20-8 = 12 \) N E; \( a = F/m = 12/5.0 = 2.4 \) m/s² east.

Q2 — Weight on Mars

An astronaut has mass \(70\) kg. On Mars, where \(g_\text{Mars} = 3.7\) m/s², what is the astronaut's weight?

Answer: N

Solution:

\( W = mg = 70 \times 3.7 = 259 \) N

Q3 — Friction on level ground

A \(15\)-kg crate is pushed horizontally with \(60\) N. Coefficient of kinetic friction \( \mu_k = 0.25 \). What is the acceleration?

Answer: m/s²

Solution:

\( N = mg = 15(9.8) = 147 \) N. \( f_k = \mu_k N = 0.25(147) = 36.75 \) N.
\( F_\text{net} = 60-36.75 = 23.25 \) N. \( a = 23.25/15 = 1.55 \) m/s²

Q4 — Frictionless incline

A block on a frictionless incline of \(25°\). What is the magnitude of acceleration down the incline?

Answer: m/s²

Solution:

\( a = g\sin\theta = 9.8 \sin(25°) = 9.8(0.4226) = 4.14 \) m/s²

Q5 — Newton's 3rd law (concept)

A horse pulls a cart forward. According to Newton's third law, the cart pulls the horse backward with equal force. Why does the horse-cart system still accelerate forward?

The horse is more massive than the cart.
The action-reaction pair acts on different objects, so they don't cancel on either body. The horse also has the ground push it forward (friction).
Newton's third law doesn't apply to living things.
The forces are not actually equal.

Solution:

The horse-on-cart and cart-on-horse forces act on different bodies — they do not cancel on the system. Acceleration is determined by the net horizontal force on each body, which includes the friction the ground exerts on the horse.

Q6 — Tension (elevator)

A \(60\)-kg person stands on a scale in an elevator accelerating upward at \(2.0\) m/s². What is the scale reading (apparent weight)?

Answer: N

Solution:

\( N - mg = ma \) (upward positive). \( N = m(g+a) = 60(9.8+2.0) = 60(11.8) = 708 \) N

Q7 — Friction on incline

A \(10\)-kg block sits on a \(20°\) incline. The static friction coefficient is \(0.40\). Will it slide?

Yes — \( mg\sin\theta > \mu_s mg\cos\theta \).
No — \( mg\sin\theta < \mu_s mg\cos\theta \) (33.5 N < 36.8 N).
Need more information about the surface.
Yes — gravity always wins on an incline.

Solution:

\( F_\parallel = mg\sin(20°) = 10(9.8)(0.342) = 33.5 \) N
\( f_s^\text{max} = \mu_s mg\cos(20°) = 0.40(10)(9.8)(0.940) = 36.8 \) N
Since \( 33.5 < 36.8 \), static friction is sufficient — the block does not slide.

Q8 — Atwood machine

Two masses, \(3.0\) kg and \(5.0\) kg, hang from a frictionless pulley. What is the acceleration of the system?

Answer: m/s²

Solution:

\( a = \frac{(m_2-m_1)g}{m_1+m_2} = \frac{(5-3)(9.8)}{8} = \frac{19.6}{8} = 2.45 \) m/s²

Q9 — Force on incline + friction

A \(6.0\)-kg sled slides down a \(30°\) snowy incline with \( \mu_k = 0.10 \). What is the acceleration?

Answer: m/s²

Solution:

\( a = g(\sin\theta - \mu_k\cos\theta) = 9.8(\sin30° - 0.10\cos30°) \)
\( = 9.8(0.500 - 0.0866) = 9.8(0.4134) = 4.05 \) m/s²

Q10 — Connected blocks (frictionless)

A 2.0-kg block and a 3.0-kg block are pushed together on a frictionless surface by a horizontal force of \(20\) N applied to the 2.0-kg block. What is the contact force between them?

Answer: N

Solution:

System: \( a = 20/(2+3) = 4.0 \) m/s²
Force on 3.0-kg block: \( F = m_2 a = 3.0(4.0) = 12 \) N