In this list you will find exercises on the main Physics topics covered in the 1st year of high school. Practice and resolve your doubts with the answers explained step by step.
Question 1 - Uniform movement (kinematics)
A car travels along a straight, deserted road and the driver maintains a constant speed of 80 km/h. After 2 hours had passed since the start of the trip, the driver drove
A) 40 km.
B) 80 km.
C) 120 km.
D) 160 km.
E) 200 km.
goal
Determine the distance traveled by the driver, in km.
Data
- The movement is uniform, that is, with constant speed and zero acceleration.
- The speed module is 80 km/h
- Travel time was 2 hours.
Resolution
Let's calculate the distance using the speed formula:
Where,
is the distance traveled in km.
is the time interval in hours.
As we want distance, we isolate in the formula.
Replacing the values:
Conclusion
When traveling at a constant speed of 80 km/h, after 2 hours of travel the driver covers 160 km.
Practice more kinematics exercises.
Question 2 - Uniformly varied movement (kinematics)
In a car race on an oval track, one of the cars is accelerating uniformly at a constant rate. The pilot starts from rest and accelerates for 10 seconds until reaching a speed of 40 m/s. The acceleration achieved by the car was
A) 4 m/s²
B) 8 m/s²
C) 16 m/s²
D) 20 m/s²
E) 40 m/s²
goal
Determine the acceleration in the 10 second time interval.
Data
10 s time interval.
Speed variation from 0 to 40 m/s.
Resolution
As there is a variation in speed, the type of movement is accelerated. Since the rate of acceleration is constant, it is a uniformly varied movement (MUV).
Acceleration is how much the speed changed over a period of time.
Where,
The is the acceleration, in m/s².
is the variation in speed, that is, final speed minus the initial speed.
is the time interval, that is, final time minus initial time.
As the car starts from rest and time begins to slow down as soon as the car starts moving, the initial speed and time are equal to zero.
Replacing the data provided in the statement:
Conclusion
In this time interval the acceleration of the car was 4 m/s².
See exercises Uniformly Varied Movement
Question 3 - Newton's First Law (dynamics)
Imagine a train that is traveling through Brazil. Suddenly, the driver has to suddenly brake the train due to an obstacle on the tracks. All objects on the train continue to move, maintaining the speed and trajectory they had before. Passengers are being thrown around the carriage, pens, books and even that apple someone brought for lunch are hovering in the air.
The principle of Physics that explains what happens inside the train carriage is
a) the Law of Gravity.
b) the Law of Action and Reaction.
c) the Law of Inertia.
d) the Energy Conservation Law.
e) the Speed Law.
Explanation
Newton's 1st law, also called the Law of Inertia, states that an object at rest will remain at rest, and an object at rest will remain at rest. An object in motion will continue to move at a constant speed unless an external force acts on it.
In this case, even with the train abruptly reducing its speed, the objects continue to move due to due to inertia, the tendency of bodies is to maintain their state of movement (direction, module and direction) or rest.
You may be interested in learning more about the Newton's First Law.
Question 4 - Newton's Second Law (dynamics)
In an experimental physics class, an experiment is carried out using boxes with different masses and applying a constant force to each one. The goal is to understand how the acceleration of an object is related to the applied force and the mass of the object.
During the experiment, the box maintains a constant acceleration of 2 m/s². Afterwards, changes in mass and strength are made in the following situations:
I - The mass is kept the same, but the force modulus is twice as large as the original.
II - The applied force is the same as the original, however, the mass is doubled.
The values of the new accelerations in relation to the original, in both cases, are, respectively
The)
B)
w)
d)
It is)
The relationship between force, mass and acceleration is described by Newton's second Law, which says: the resultant force acting on a body is equal to the product of its mass and its acceleration.
Where,
FR is the resultant force, the sum of all forces acting on the body,
m is the mass,
a is the acceleration.
In situation I, we have:
The mass remains the same, but the magnitude of the force is doubled.
To differentiate, we use 1 for the original quantities and 2 for the new one.
Original:
New:
Force 2 is double force 1.
F2 = 2F1
As the masses are equal, we isolate them in both equations, equate them and solve for a2.
Replacing F2,
Thus, when we double the magnitude of the force, the magnitude of the acceleration is also multiplied by 2.
In situation II:
Equalizing the forces and repeating the previous process:
Replacing m2,
Thus, by doubling the mass and maintaining the original force, the acceleration drops by half.
Need reinforcement with Newton's Second Law? Read our content.
Question 5 - Newton's Third Law (dynamics)
A physics teacher, excited about practical learning, decides to carry out a peculiar experiment in the classroom. He puts on a pair of roller skates and then pushes against a wall. We will explore the physical concepts involved in this situation.
When pushing against the classroom wall while wearing a pair of roller skates, what will happen to the teacher and what are the physical concepts involved?
a) A) The teacher will be projected forward, due to the force applied to the wall. (Newton's Law - Third Law of Action and Reaction)
b) The teacher will remain still, as there is friction between the skates and the floor. (Newton's Law - Conservation of Quantity of Linear Motion)
c) The teacher remains still. (Newton's Law - Friction)
d) The teacher will be thrown backwards, due to the rolling of the skates, due to the application of the wall reaction. (Newton's Law - Third Law of Action and Reaction)
e) The teacher's skates will heat up due to friction with the floor. (Newton's Law - Friction)
Newton's third law explains that every action produces a reaction of the same intensity, same direction and opposite direction.
When applying a force against the wall, the reaction pushes the teacher in the opposite direction, with the same intensity as the applied force.
The Law of action and reaction acts on pairs of bodies, never on the same body.
As the skates allow rolling, the teacher's center of mass is thrown backwards and he slides across the room.
Remember the Newton's Third Law.
Question 6 - Law of universal gravitation
The school's Physics club is exploring the Moon's orbit around the Earth. They wish to understand the force of gravitational attraction between the Earth and its natural satellite, applying the principles of Newton's Law of Universal Gravitation.
Mass estimates are kg for Earth and about 80 times smaller for the Moon. Their centers are located at an average distance of 384,000 km.
Knowing that the constant of universal gravitation (G) is N⋅m²/kg², the force of gravitational attraction between the Earth and the Moon is approximately
The)
B)
w)
d)
It is)
Newton's Law of Universal Gravitation says that: "The force of gravitational attraction between two masses (m1 and m2) is directly proportional to the product of their masses and the universal constant of gravitation and inversely proportional to the square of two distance.
Its formula:
where:
F is the force of gravitational attraction,
G is the constant of universal gravitation,
m1 and m2 are the masses of the bodies,
d is the distance between the centers of the masses, in meters.
Value Replacement:
See more about Gravitational Force.
Question 7 - Free fall (Movement in a uniform gravitational field)
In a practical assignment for the school's Science Fair, a group will expose the effects of a uniform gravitational field. After an explanation of the concept of gravity, they perform a practical experiment.
Two steel spheres, one with a diameter of 5 cm and the other with a diameter of 10 cm, are released from rest, in the same moment, by one of the group members, from a window on the third floor of the school.
On the ground, a cell phone that records in slow motion records the exact moment of impact of the spheres on the ground. On a sheet, the group asks spectators to select the option that, according to them, explains the relationship between the speeds of objects when they touch the ground.
You, with a good understanding of Physics, will select the option that says
a) the heavier object will have a greater speed.
b) the lighter object will have a greater speed.
c) both objects will have the same speed.
d) the difference in speed depends on the height of the tower.
e) the difference in speed depends on the mass of the objects.
Neglecting the effects of air, all objects fall with the same acceleration due to gravity, regardless of their mass.
The gravitational field attracts objects to the center of the Earth with the same constant acceleration of approximately .
The speed function is described by:
With Vi being the initial velocity equal to zero and the acceleration being g:
The speed, therefore, only depends on the value of the acceleration due to gravity and the time of fall.
Distance traveled can also be measured by:
It is possible to see that neither the speed nor the distance depend on the mass of the object.
Train more free fall exercises.
Question 8 - Horizontal launch (Movement in a uniform gravitational field)
A pair of students, in an experiment, throw a ball horizontally from a high height. While one throws the ball, the other at a given distance records a video of the ball's trajectory. Neglecting air resistance, the trajectory and horizontal speed of the ball during movement are
a) a straight descending line, and the horizontal speed will increase.
b) a straight line, and the horizontal speed will increase with time.
c) an arc of a circle, and the horizontal speed will decrease with time.
d) a wavy line, and the horizontal speed will fluctuate.
e) a parabola, and the horizontal velocity will remain constant.
Horizontal and vertical movement are independent.
When air resistance is ignored, the horizontal speed will be constant, since there is no friction, and the movement is uniform.
Vertical movement is accelerated and depends on the acceleration of gravity.
The composition of the movements forms the trajectory of a parabola.
Are you interested in learning more about Horizontal Launch.
Question 9 - Power and performance
A student is investigating the efficiency of a machine which, according to the manufacturer's information, is 80%. The machine receives a power of 10.0 kW. Under these conditions, the useful power offered and the power dissipated by the machine are, respectively
a) useful power: 6.4 kW and dissipated power: 3.6 kW.
b) useful power: 2.0 kW and dissipated power: 8.0 kW.
c) useful power: 10.0 kW and dissipated power: 0.0 kW.
d) useful power: 8.0 kW and dissipated power: 2.0 kW.
e) useful power: 5.0 kW and dissipated power: 5.0 kW.
Efficiency (η) is the ratio between useful power and received power, expressed as:
Useful power, in turn, is the power received minus the power dissipated.
Useful power = received power - dissipated power
With the yield being 80%, or 0.8, we have:
Thus, the useful power is:
Useful power = received power - dissipated power
Useful power = 10 kW - 2 W = 8 kW
You might want to remember about mechanical power and performance.
Question 10 - Conservative mechanical system
In a Physics laboratory, a track with carts simulates a roller coaster. They abandon the cart from rest at the highest point of the trail. The cart then descends, decreasing its height, while its speed increases during the descent.
If there is no energy loss due to friction or air resistance, how does conservation of mechanical energy apply to this conservative system?
a) Total mechanical energy increases as the cart is gaining speed.
b) The total mechanical energy decreases, as part of the energy is converted into heat due to friction.
c) The total mechanical energy remains constant, as there are no dissipative forces acting.
d) The total mechanical energy depends on the mass of the cart, as it affects the gravitational force.
e) The total mechanical energy varies depending on the ambient temperature, as it affects air resistance.
Mechanical energy is the sum of its parts, such as gravitational potential energy and kinetic energy.
Considering the conservative system, that is, without energy losses, the final energy must be equal to the initial one.
At the beginning, the cart was stationary, with its kinetic energy equal to zero, while its potential energy was the maximum, as it was at the highest point.
When descending, it begins to move and its kinetic energy increases as the height decreases, also decreasing its potential energy.
While one portion decreases, the other increases in the same proportion, keeping the mechanical energy constant.
Remember the concepts about mechanical energy.
Question 11 - Specific mass or absolute density
In an investigation into the properties of matter, three cubes of different volumes and materials are used to create a scale of the specific mass of these materials.
With the help of a scale and a ruler, the following are obtained for the cubes:
- Steel: Mass = 500 g, Volume = 80 cm³
- Wooden: Mass = 300 g, Volume = 400 cm³
- Aluminum: Mass = 270 g, Volume = 100 cm³
From the highest specific mass to the lowest, the values found are:
a) Steel: 6.25 g/cm³, Aluminum: 2.7 g/cm³, Wood: 0.75 g/cm³
b) Wood: 1.25 g/cm³, Steel: 0.75 g/cm³, Aluminum: 0.5 g/cm³
c) Steel: 2 g/cm³, Wood: 1.25 g/cm³, Aluminum: 0.5 g/cm³
d) Aluminum: 2 g/cm³, Steel: 0.75 g/cm³, Wood: 0.5 g/cm³
e) Aluminum: 2 g/cm³, Steel: 1.25 g/cm³, Wood: 0.75 g/cm³
The specific mass of a material is defined as the mass per unit volume, and is calculated by the formula:
For the steel:
To the wood:
For the aluminum:
Learn more at:
- Especific mass
- Density
Question 12 - Pressure exerted by a liquid column
A student is diving into a lake at sea level and reaches a depth of 2 meters. What is the pressure that the water exerts on it at this depth? Consider the acceleration due to gravity as and the density of water as
.
a) 21 Pa
b) 121 Pa
c) 1121 Pa
d) 121,000 Pa
e) 200,000 Pa
The pressure in a fluid at rest is given by the formula:
P=ρ⋅g⋅h + atmospheric P
where:
P is the pressure,
ρ is the density of the fluid,
g is the acceleration due to gravity,
h is the depth of the fluid.
Practice more hydrostatic exercises.
ASTH, Rafael. Physics exercises (solved) for 1st year of high school.All Matter, [n.d.]. Available in: https://www.todamateria.com.br/exercicios-de-fisica-para-1-ano-do-ensino-medio/. Access at:
See too
- Exercises on potential and kinetic energy
- Physics Formulas
- Newton's Laws Exercises commented and solved
- Work in Physics
- Hydrostatic exercises
- Physics at Enem
- Exercises on kinetic energy
- Gravity
