Multiple Choice Quiz (Practice) Here are 10 multiple-choice questions covering all the concepts we just reviewed. Try to answer them without looking at the solutions first. Newton's Second Law states that: a) A body in motion stays in motion. b) Force is equal to mass times velocity. c) Force is equal to mass times acceleration. d) For every action, there is an equal and opposite reaction. A 50-kg crate weighs approximately: a) 5 N b) 50 N c) 500 N d) 5000 N What does the Normal Force always do in a Work-Energy problem? a) Positive work. b) Negative work. c) No work. d) It depends on the angle. When solving for acceleration using Force & Acceleration, the FIRST step should always be to: a) Apply the kinematic equations. b) Draw a Free Body Diagram. c) Write T 1 + Σ U 1 − 2 = T 2 T 1 ​ +ΣU 1−2 ​ =T 2 ​ . d) Find the normal force. An object is moving to the right. If friction is acting on the object, the work done by friction is: a) Positive. b) Negative. c) Zero. d) Cannot be determined. The equation for Kinetic Energy is: a) T = m v T=mv b) T = 1 2 m v 2 T= 2 1 ​ mv 2 c) T = m g h T=mgh d) T = 1 2 k x 2 T= 2 1 ​ kx 2 A 2-kg ball is thrown straight up. As it moves upward, the work done by its weight (gravity) is: a) Positive. b) Negative. c) Zero. d) Increasing. A car accelerates from rest to 20 m/s. Its final kinetic energy is 40,000 J. What is its mass? a) 100 kg b) 200 kg c) 1000 kg d) 2000 kg A 5-kg box is pulled horizontally with a force of 30 N over a distance of 10 m. Friction is neglected. What is the final speed of the box if it starts from rest? a) 7.75 m/s b) 10.95 m/s c) 15.49 m/s d) 60 m/s An object has a mass of 10 kg and is accelerating at 4 m/s². What is the net force acting on it? a) 2.5 N b) 14 N c) 40 N d) 400 N Answer Key and Explanations F = m a F=ma W = m g = 50 ( 81 = 5 N W=mg=50(9.81)=490.5N, which is approximately 500 N. No work. Work is F ⋅ d ⋅ cos ⁡ ( θ F⋅d⋅cos(θ), and θ = 90 ° θ=90° for a normal force. Draw a Free Body Diagram. This is the key to visualizing all the forces. Negative work. Friction opposes the motion. Kinetic energy is T = 1 2 m v 2 T= 2 1 ​ mv 2 . Negative work. The force (gravity) acts downward, while the displacement is upward. 40000 = 1 2 ( m ( 20 2 → 40000 = 200 m → m = 200 k g 40000= 2 1 ​ (20) 2 →40000=200m→m=200kg. T 1 = 0 T 1 ​ =0, U = F ⋅ d = 30 ( 10 = 300 J U=F⋅d=30(10)=300J. 300 = 1 2 ( 5 v 2 → 300 = 5 v 2 → v 2 = 120 → v = 95 m / s 300= 2 1 ​ (5)v 2 →300=2.5v 2 →v 2 =120→v=10.95m/s. F = m a = 10 ( 4 = 40 N F=ma=10(4)=40N.

Weight ($W$) is the force of gravity acting on an object, calculated as mass ($m$) multiplied by the acceleration due to gravity ($g$). We use $g \approx 9.81 \text{ m/s}^2$.

$$W = mg$$

$$W = (50 \text{ kg})(9.81 \text{ m/s}^2)$$ $$W = 490.5 \text{ N}$$

This value is approximately 500 N.

---

3. What does the Normal Force always do in a Work-Energy problem?

c) No work.

Explanation:
Work done by a force is given by $W = Fd \cos(\theta)$, where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and displacement vectors. The normal force always acts perpendicular to the surface of contact, and thus perpendicular to the displacement if the object moves along that surface. When $\theta = 90^\circ$, $\cos(90^\circ) = 0$, so the work done by the normal force is zero.

---

4. When solving for acceleration using Force & Acceleration, the FIRST step should always be to:

b) Draw a Free Body Diagram.

Explanation:
A Free Body Diagram (FBD) is essential for visualizing all the forces acting on an object. By drawing an FBD, you can correctly identify and resolve all forces, which is crucial before applying Newton's Second Law ($\Sigma F = ma$) to solve for acceleration.

---

5. An object is moving to the right. If friction is acting on the object, the work done by friction is:

b) Negative.

Explanation:
Friction is a force that always opposes the direction of motion. If an object is moving to the right, the friction force acts to the left. Since the force of friction and the displacement are in opposite directions (angle $\theta = 180^\circ$), the work done by friction ($W = Fd \cos(180^\circ)$) will be negative.

---

6. The equation for Kinetic Energy is:

b) $T=\frac{1}{2}mv^2$

Explanation:
Kinetic energy ($T$) is the energy an object possesses due to its motion. It is defined by the formula:

$$T = \frac{1}{2}mv^2$$

where $m$ is the mass of the object and $v$ is its speed.

---

7. A 2-kg ball is thrown straight up. As it moves upward, the work done by its weight (gravity) is:

b) Negative.

Explanation:
The weight (force of gravity) on the ball always acts downward. As the ball moves upward, its displacement is in the opposite direction to the force of gravity. Therefore, the angle between the force and displacement is $180^\circ$, resulting in negative work done by gravity.

---

8. A car accelerates from rest to 20 m/s. Its final kinetic energy is 40,000 J. What is its mass?

b) 200 kg

We are given $T = 40,000 \text{ J}$ and $v = 20 \text{ m/s}$. We need to find $m$.

$$40,000 \text{ J} = \frac{1}{2} m (20 \text{ m/s})^2$$

$$40,000 = \frac{1}{2} m (400)$$ $$40,000 = 200m$$ $$m = \frac{40,000}{200}$$ $$m = \boxed{200 \text{ kg}}$$

---

9. A 5-kg box is pulled horizontally with a force of 30 N over a distance of 10 m. Friction is neglected. What is the final speed of the box if it starts from rest?

b) 10.95 m/s

Since friction is neglected and the force is horizontal, the work done ($W$) is simply the force ($F$) multiplied by the distance ($d$).

$$W = Fd$$ $$W = (30 \text{ N})(10 \text{ m})$$ $$W = 300 \text{ J}$$

The Work-Energy Theorem states that the net work done on an object equals the change in its kinetic energy ($\Delta T$). Since the box starts from rest, its initial kinetic energy ($T_i$) is 0.

$$W_{net} = \Delta T = T_f - T_i$$ $$300 \text{ J} = T_f - 0$$ $$T_f = 300 \text{ J}$$

$$T_f = \frac{1}{2}mv_f^2$$ $$300 \text{ J} = \frac{1}{2} (5 \text{ kg}) v_f^2$$ $$300 = 2.5 v_f^2$$ $$v_f^2 = \frac{300}{2.5}$$ $$v_f^2 = 120$$ $$v_f = \sqrt{120}$$ $$v_f \approx \boxed{10.95 \text{ m/s}}$$

---

10. An object has a mass of 10 kg and is accelerating at 4 m/s². What is the net force acting on it?

c) 40 N

Newton's Second Law states that the net force ($\Sigma F$) acting on an object is equal to its mass ($m$) multiplied by its acceleration ($a$).

$$\Sigma F = ma$$

$$\Sigma F = (10 \text{ kg})(4 \text{ m/s}^2)$$ $$\Sigma F = \boxed{40 \text{ N}}$$