Newton's laws exercises

A force represents a directed interaction.

The system is the door and the external agent is the student.

The relevant force is named student on door.

The classification asks whether macroscopic contact exists.

Gravity acts between Earth and the ball without surface contact.

The other choices describe contact.

Net force is the vector sum, not a third interaction.

ΣFₓ = +18 N − 11 N.

ΣFₓ = +7 N.

The resultant points toward +x.

Vector addition is performed by components.

ΣFₓ = 6 − 1 and ΣFᵧ = 2 + 5.

ΣF⃗ = (5, 7) N.

Both components are positive in the chosen axes.

Only the two solid forces belong to the interaction inventory.

The dashed arrow summarizes their vector sum.

It is not added to the FBD as an independent force.

The first law connects zero net force with constant velocity.

Here that constant velocity is zero.

Individual balanced forces may still exist.

The velocity vector does not change.

Therefore a = 0.

In an inertial frame, $\sum\vec F=0$ .

The individual forces exist and are opposite.

Their vector sum is zero.

Acceleration is zero; the box may be at rest or moving at constant velocity.

Inertia preserves velocity when $\sum\vec F=0$ .

A net force produces acceleration, which is a change in velocity.

Motion alone does not prove a forward net force.

The first law applies in the stated inertial frame.

Zero net force implies zero acceleration.

Initial velocity may be anything, but it remains constant.

For constant mass, $\sum F_x=ma_x$ .

The +x axis agrees with the resultant.

aₓ = 18 N / 4 kg = 4.5 $m/s^2$ .

Acceleration points toward +x.

Net force and acceleration are parallel.

$\sum\vec F=m\vec a$ .

|ΣF| = (6 kg)(3 $m/s^2$ ) = 18 N.

The resultant points west.

The relation refers to the same system.

m = F_net/a.

m = 24 N /(6 $m/s^2$ ) = 4 kg.

The positive mass measures its inertia.

The vector equation is projected onto x and y.

aₓ = 8/2 and aᵧ = −6/2.

a⃗ = (4, −3) $m/s^2$ .

Signs come from the chosen axes.

At fixed mass, a is linear in F_net.

The slope-intercept form has slope 1/m.

For m = 2 kg, 1/m = 0.5 kg⁻¹.

Both bodies receive the same net force.

a₃/a₉ = (F/3)/(F/9) = 3.

The lower-mass body accelerates three times as much.

Net force determines acceleration.

Both point north.

Instantaneous velocity may keep an eastward component while beginning to change.

External net force connects with the rate of change of momentum.

$\sum\vec F_{ext}=m\vec a$ is the form used here for constant mass.

A variable-mass system must model matter flow before simplifying.

Mass is the factor relating net force and acceleration.

For the same resultant, greater mass means smaller acceleration.

It therefore measures inertia and is expressed in kg.

Weight is the local gravitational force.

$W=mg$ .

W = (7.5 kg)( $9.8\,m/s^2$ ) = 73.5 N.

It points toward Earth's centre.

The 49 N datum is a force.

m = W/g.

m = 49 N /( $9.8\,m/s^2$ ) = 5 kg.

The answer is mass and is expressed in kg.

The local model is $W=mg$ .

W₁ = 10(9.8) and W₂ = 10(1.6).

W₁ = 98 N and W₂ = 16 N.

Weight changes; mass remains 10 kg.

Mass is a scalar and its SI unit is kg.

Weight is a gravitational force.

Weight is therefore expressed in N.

The relevant interaction is table–book.

One force is table on book.

Its partner is book on table and acts on the table.

Person and cart form an interaction pair.

The forces appear simultaneously and are opposite.

Their magnitudes are equal although accelerations may differ.

Both forces belong to the same collision.

The third law requires simultaneity, equal magnitude, and opposition.

Different responses arise from different masses and structures.

The person exerts a force on the wall.

The wall simultaneously exerts an opposite force on the person.

That second force produces the sensation in the hands.

The mutual pair has magnitude 150 N on each skater.

For the 50 kg system, horizontal net force is 150 N.

a = 150/50 = 3 $m/s^2$ .

The other skater has a different acceleration because their mass is 75 kg.

The third law relates forces in one interaction on different receivers.

A balance adds forces sharing the same receiving system.

Only the latter can cancel within one body's FBD.

The isolated system is the box.

Support, weight, and push are forces exerted on it.

The diagram excludes kinematic quantities and forces on other bodies.

Velocity describes the state of motion.

It has no agent and receiver as a force does.

It must remain outside the FBD.

The FBD contains forces acting on the box.

The box-on-hand force has the hand as receiver.

It belongs to the hand's FBD, not the box's.

The system is the lamp.

Earth exerts gravity and the cord exerts a contact interaction.

The ceiling interacts with the cord, not directly with the lamp in this model.

Both forces act on the same system.

ΣFₓ = +7 N − 3 N.

ΣFₓ = +4 N.

The resultant points right.

The components project one force.

Fₓ = F cos θ and Fᵧ = F sin θ.

Fₓ = 8 N and Fᵧ = 6 N.

Do not add them to vector F in the force inventory.

The FBD fixes the forces acting on the box.

ΣFₓ = 9 − 3 = 6 N.

aₓ = ΣFₓ/m = 6/2 = 3 $m/s^2$ .

Acceleration points toward +x.

S is inertial by hypothesis.

Relative velocity V is constant.

Acceleration is preserved and S' belongs to the same class of inertial frames.

Define +x along the train's motion.

v_person/S = v_person/S' + V_S'/S.

v = 3 + 12 = 15 m/s.

The frames assign different velocities to the same motion.

Relative velocity V does not change.

Differentiating gives dV/dt = 0.

Therefore $\vec a'=\vec a$ .

An inertial frame may translate at constant velocity.

The car increases speed and therefore accelerates.

That frame does not preserve standard form directly without corrections.

A free object should preserve velocity in an inertial frame.

A common unexplained acceleration suggests motion of the frame.

Check the frame before assigning physical forces to the objects.