Physics Study Guide for the HESI Exam

General Information

Not every student will need to take the Physics section of the HESI. It depends on the type of nursing program to which you are applying. If your chosen program requires knowledge of how matter reacts (such as in medical imaging), you will probably need to take the Physics section. Check with your test provider to determine whether to study for this section of the HESI.

If you do need to sit for the Physics section of the HESI exam, you will have 25 minutes to complete 25 questions. This guide will review the most important physics topics that you’ll need to know.

Motion

Motion simply means a change in position over time. Anytime something moves, from a car driving down a street to a person walking across a room, we say that the object is in motion. In science, we describe motion precisely using measurements like speed, velocity, and acceleration to understand how things move.

The Nature of Motion

Motion is everywhere. From the Earth rotating on its axis to a ball rolling across the floor, motion happens when an object changes its position relative to something else. We usually describe motion in terms of how fast something moves (speed), the direction it moves (velocity), and how its movement changes (acceleration). This allows us to break down motion into measurable parts.

Speed

Speed indicates how quickly something is moving. It is a scalar quantity, meaning it tells you how fast but not the direction.

We calculate speed (\(s\)) by dividing distance (\(d\)) by time (\(t\)). As a formula, this is represented as:

\[s=d \div t \text{ or } s=\frac{d}{t}\]

We can also calculate the average speed, which is helpful when the speed changes over time:

\[\overline{s}=\frac{d_t}{t_t}\]

Let’s try an example problem.

A nurse walks \(60\) meters in \(30\) seconds while making her rounds. What is her average speed?

Solution

We can see that the total distance is \(60\) meters and the total time is \(30\) seconds, which makes calculating the average speed straightforward:

\[\overline{s} = \frac{d_t}{t_t}= \frac{60 \text{ m}}{\text{30 s}} = 2 \text{ m/s}\]

So, the nurse’s average speed is two meters per second.

Velocity

Velocity is similar to speed, but with a key difference: It includes direction. Because of that, it’s a vector quantity, which means it has both magnitude (how fast) and direction (e.g., north, east, or downward). That’s what sets it apart from speed, which is scalar.

The formula for velocity (\(v\)) is similar to speed, except that instead of distance we use displacement (\(D\)):

\[v=\frac{D}{t}\]

Displacement refers to the change in position from the starting point to the ending point, measured along a straight line, not the total path traveled. If a person is pacing back and forth in a straight line, their average speed could be two meters per second, but their overall velocity would be zero.

We’ll do an example problem.

A patient is wheeled \(10\) meters north down a hallway, then \(5\) meters south, all in \(10\) seconds. What was their velocity?

Solution

Since we’re finding velocity, we need to find the patient’s displacement rather than distance:

\[v = \frac{10 \text{ m} - 5 \text{ m}}{10 \text{ s}} = \frac{5 \text{ m}}{10 \text{ s}} = 0.5 \text{ m/s north}\]

So, the patient’s velocity is \(0.5\) meters per second to the north.

Acceleration

Acceleration is the rate of change of velocity of an object. It is measured in meters per second squared (\(m/s^{2}\)). As a formula, it is:

\[a = \frac{\Delta v}{t}\]

where \(\Delta v\) is the change of velocity and \(t\) is the time.

For instance, if a car increases its speed from \(10 \text{ m/s}\) to \(20 \text{ m/s}\) in \(20\) seconds, its acceleration will be:

\[\frac{20-10}{20}=\frac{10}{20}= 0.5 \text{ m/s}^{2}\]

Note: If there is deceleration (negative acceleration), the answer will include a negative sign.

Here is an example problem.

A car slows down from \(20\) to \(0\) meters per seconds in \(10\) seconds. What is its acceleration?

\[a = \frac{\Delta v}{t} = \frac{0 \text{ m/s} - 20 \text{ m/s}}{10 \text{ s}} = \frac{-20 \text{ m/s}}{10 \text{ s}} = -2 \text{m/s}^2\]

So the car’s acceleration is \(–2 \text{ m/s}^2\). Remember, the negative sign means that the car is slowing down.

Projectile Motion

Projectile motion is the curved path something follows when launched into the air, and it is only affected by gravity. Think of a soccer ball being kicked or a basketball shot into the hoop.

Projectile motion actually involves two motions happening at once:

horizontal motion—motion at constant speed in the \(x\)-axis (horizontal plane)
vertical motion—accelerated motion in the \(y\)-axis (vertical plane) due to gravity (usually \(g = -9.8 \text{ m/s}^2\))

Note: When \(g\) refers to the downward pull of gravity, it is given as a negative value. When it simply refers to the magnitude of the force, it will be given as a positive value. For simplicity, you will often use \(g = 10\text{ m/s}^2\), which will give you an approximate value that is close enough for most problems you’ll encounter.

Therefore, to determine projectile motion, you must work out each of these motions separately.

Horizontal motion is described by this equation:

\[x = x_0 + v_x \cdot t\]

where \(x\) and \(x_0\) are the final and initial positions, respectively, \(v_x\) is the velocity in the \(x\)-axis, and \(t\) is the time. To determine the velocity in the \(x\)-axis for a projectile thrown upward, you need to know its velocity (\(v\)) and at what angle it was thrown in respect to the horizontal (\(\theta\)). This is the formula for that:

\[v_{x} = v \times \cos(\theta)\]

Vertical motion is described by these equations:

\[y = y_0 + v_{0y} \cdot t + \frac{1}{2} a \cdot t^2\] \[v_y = v_{0y} + a \cdot t\] \[v_y^2 = v_{0y}^2 + 2a(y - y_0)\]

where \(y\) and \(y_0\) are the final and initial positions in the \(y\)-axis, respectively, \(v_y\) and \(v_{0y}\) are the final and initial velocities in the \(y\)-axis, \(a\) is the acceleration (usually \(a = g = -9.8 \text{ m/s}^2\)), and \(t\) is the time.

As with horizontal motion, you need to be able to determine the velocity in the \(y\)-axis for a projectile thrown upward. To do that, you multiply its velocity (\(v\)) by the sine of its angle with respect to the horizontal (\(\theta\)):

\[v_y = v \cdot \sin(\theta)\]

This concept can seem tricky, so let’s do an example problem to see how this all works.

A ball is rolled off the edge of a table that is 20 meters high with an initial horizontal velocity of five meters per second. Assume no air resistance and use \(g = 10\text{ m/s}^2\).

a) How long does it take for the ball to reach the ground?

b) How far from the base of the table does the ball land?

c) What is the horizontal velocity just before it hits the ground?

d) What is the vertical velocity just before it hits the ground?

When we work with projectile motion, it helps to remember that horizontal and vertical motions are independent. The horizontal direction has constant velocity (no acceleration), while the vertical direction has constant acceleration due to gravity. Let’s carefully solve this one step at a time.

a) Time to ground: We know the ball starts with no vertical velocity, so its vertical displacement (\(y\)) comes only from gravity. As such, we use this formula from above:

\[y = y_0 + v_{0y} \cdot t + \frac{1}{2} a \cdot t^2\]

However, we can simplify it to this:

\[y = \frac{1}{2}a \cdot t^2\]

Substituting, \(y = 20\text{ m}\) and \(a = 10\text{ m/s}^2\), we get:

\[20 = \frac{1}{2}(10)t^2\] \[20 = 5t^2\] \[4 = t^2\] \[t = 2\]

This tells us the ball is in the air for \(2\) seconds.

b) Horizontal distance traveled: Since horizontal velocity stays constant, we can use:

\[x = v_x \cdot t\]

where \(x\) is the horizontal distance and \(v_x\) is the velocity in the \(x\)-axis.

Substituting \(v_x = 5\text{ m/s}\) and \(t = 2\text{ s}\), we get:

\[x = 5 \cdot 2 = 10\]

So, the ball lands 10 meters away from the base of the table.

c) Horizontal velocity at impact: There is no horizontal acceleration, so the horizontal velocity at impact is exactly the same as the initial horizontal velocity: \(5 \text{ m/s}\).

d) Vertical velocity at impact: Vertical velocity increases steadily because of gravity, according to the equation:

\[v_y = v_{0y} + a \cdot t\]

So:

\[v_y = 0 + (10)(2) = 20\]

The velocity of \(20\text{ m/s}\) is directed downward.

Putting it all together, the ball takes two seconds to hit the ground, lands \(10\) meters away, keeps its horizontal velocity of five meters per second, and gains a downward velocity of \(20\) meters per second at impact.

Newton’s Laws of Motion

Sir Isaac Newton developed three fundamental laws that describe how objects move. They relate to the concept of force, which is any push or pull acting on an object. It is a vector, so it has magnitude and direction, and its International System of Units (SI) unit is the newton (\(\text{N}\)).

An unbalanced force occurs when one force acting on an object is stronger than another force in the opposite direction. This causes a change in motion.
A net force is the total force acting on an object. If all the forces are balanced (canceled out), the net force is zero, and there is no change in motion.

Newton’s First Law

Newton’s first law states that an object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by an unbalanced external force.

Put simply, this means if nothing pushes or pulls on an object, it won’t change what it is doing. That is the concept of inertia, the resistance of an object to changing its motion.

There is no specific formula for Newton’s first law, but the idea of balanced and unbalanced forces is key.

Newton’s Second Law

Newton’s second law states that the acceleration of an object depends on the net force acting on it and its mass. The more force you apply, the faster it accelerates. But heavier objects need more force to accelerate.

The formula for this law is:

\[F = m \cdot a\]

where \(F\) is the force, \(m\) is the mass, and \(a\) is the acceleration.

The force is measured using newtons. A newton is the force required to accelerate a mass of one kilogram at a rate of one meter per second squared.

Here are a couple example problems

A \(5\text{ kg}\) box is pushed with a force of \(20\text{ N}\). What is its acceleration?

Solution

Since we have the mass and the force and want to know acceleration, we simply reconfigure the above formula to get our answer:

\[a = \frac{F}{m} = \frac{20 \text{ N}}{5 \text{ kg}} = 4 \text{ m/s}^2\]

Note: Remember from above that acceleration is measured in meters per second squared.

How much force is needed to make a \(2\text{ kg}\) object accelerate at \(3 \text{ m/s}^2\)?

Solution

\[F = ma = 2 \text{ kg} \times 3 \text{ m/s}^2 = 6 \text{ N}\]

Newton’s Third Law

This law states that for every action there is an equal and opposite reaction. This means that if object A exerts a force on object B, object B exerts the same force back on object A but in the opposite direction. For example, when you jump off a boat, you push the boat backward (action), and the boat pushes you forward (reaction). This is why the boat moves when you leap.

Friction

Friction is the force that resists motion when two surfaces touch. It always acts in the opposite direction of motion. Friction is what makes slip-resistance shoes effective or what allows two sticks rubbing together to create fire.

There are two main types of friction:

static friction—keeps an object at rest from moving
kinetic friction—slows down a moving object

The formula to find the friction force (\(F_f\)) is:

\[F_f = \mu \cdot N\]

where \(\mu\) is the coefficient of friction and \(N\) is the normal force pressing the object into the surface. For static friction, use the coefficient of static friction (\(\mu_s\)), and for kinetic friction, use the coefficient of kinetic friction (\(\mu_k\)).

The following diagram shows the different forces acting on an object. Note each of these forces:

\(N\) is the normal force.
\(w\) is the weight of the object (equal to \(\text{mg}\) in this diagram).
\(F\) is the horizontal force applied to the object.
\(f_s\) and \(f_k\) are the static and kinetic friction forces, respectively.

The diagram labeled (a) shows a static object that has static friction acting on it, and the diagram labeled (b) shows a moving object that has kinetic friction acting on it.

1 Friction and Forces.jpeg

Retrieved from: https://openstax.org/books/university-physics-volume-1/pages/6-2-friction

We’ll do an example problem.

A \(10\text{ kg}\) box is sliding across the floor. You apply a force of \(50 \text{ N}\) to the right. The coefficient of kinetic friction is \(0.2\). What is the net force on the box? (Use \(g = 10\text{ m/s}^2\))

Solution

We start by calculating the normal force (\(N\)) on the box. It is equal to the weight (mass times gravity) because the box is on a flat surface, so:

\[N = w = mg = (10)(10) = 100 \text{ N}\]

Now, we calculate the friction force (\(F_f\)). Since the box is moving, this is kinetic friction (\(F_k\)):

\[F_k = \mu_kN = (0.2)(100) = 20 \text{ N}\]

Then, we find the net force on the box by subtracting the kinetic friction force from the applied force:

\[F_n = F - F_k = 50 \text{ N} - 20 \text{ N} = 30 \text{ N}\]

So, the box accelerates to the right with a net force of \(30 \text{ N}\).