Practice · Unit 3

Forces and equations of motion exercises

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Equilibrium

Exercise to explore

Equilibrium while moving

Type
Conceptual
Difficulty
1/5
Time
5 min

A probe moves in a straight line at a constant 12 m/s relative to an inertial frame. Which statement is correct?

Request a hint
  • Equilibrium concerns change in velocity, not zero velocity.
Review the solution
  1. Principle

    The velocity vector is constant.

  2. Representation

    Therefore a=0.

  3. Calculation

    In an inertial frame, ΣF_ext=0.

Equilibrium

Exercise to explore

Unknown force for equilibrium

Type
Numerical
Difficulty
2/5
Time
5 min

Forces F1=(8,-3) N, F2=(-2,5) N and an unknown F3 act on a ring. Find the components of F3 for equilibrium.

Request a hint
  • At equilibrium, the sum of each component is zero.
Review the solution
  1. Principle

    ΣF_x=8-2+F3x=0.

  2. Representation

    F3x=-6 N.

  3. Calculation

    ΣF_y=-3+5+F3y=0.

  4. Interpretation

    F3y=-2 N; F3 cancels the resultant of F1+F2.

Equilibrium

Exercise to explore

Two symmetric cables

Type
Numerical
Difficulty
2/5
Time
5 min

A 120 N lamp is in equilibrium, supported by two identical cables 35° above the horizontal. Find the tension in each cable.

A central lamp has two symmetric angled tensions and its downward weight.left Tright Tweight

Tensions follow the cables. Their horizontal components cancel and their vertical components support the weight, without revealing T numerically.

Request a hint
  • The vertical components of both tensions must add to 120 N.
Review the solution
  1. Principle

    Horizontal components cancel by symmetry.

  2. Representation

    ΣF_y=2T sin35°-120=0.

  3. Calculation

    T=120/(2sin35°)≈104.6 N.

  4. Interpretation

    The two cables share the vertical support component.

Equilibrium

Exercise to explore

Parallel force holding a block

Type
Numerical
Difficulty
3/5
Time
5 min

A 5.0 kg block rests on a frictionless 30° incline. What force parallel to the plane and up the slope keeps it in equilibrium? Use g=9.8 m/s².

Request a hint
  • Balance the component of weight parallel to the plane.
Review the solution
  1. Principle

    The downslope weight component is mg sin30°.

  2. Representation

    mg sin30°=(5.0)(9.8)(0.5)=24.5 N.

  3. Calculation

    The required applied force is 24.5 N upslope.

Dynamics

Exercise to explore

Acceleration of two blocks

Type
Numerical
Difficulty
2/5
Time
5 min

Two 2 kg and 3 kg blocks touch on a frictionless horizontal surface. A 20 N external force pushes the 2 kg block toward the 3 kg block. Find their common acceleration.

Two touching blocks lie inside a combined boundary; another boundary isolates the second block while an external force pushes the first.system 1+2system 2m₁m₂external Fcontact on 2

The small boundary isolates block 2 and shows contact. The large boundary includes both blocks, so contact is internal and leaves the external balance.

Request a hint
  • Treat both blocks as one system.
Review the solution
  1. Principle

    Choose both blocks as the system.

  2. Representation

    Total mass is 5 kg and external horizontal force is 20 N.

  3. Calculation

    a=20/5=4 m/s².

  4. Interpretation

    Contact force is internal to this system.

Dynamics

Exercise to explore

Internal contact force

Type
Numerical
Difficulty
3/5
Time
5 min

For the 2 kg and 3 kg blocks accelerated by 20 N on a frictionless surface, find the contact force exerted on the 3 kg block.

Two touching blocks lie inside a combined boundary; another boundary isolates the second block while an external force pushes the first.system 1+2system 2m₁m₂external Fcontact on 2

The small boundary isolates block 2 and shows contact. The large boundary includes both blocks, so contact is internal and leaves the external balance.

Request a hint
  • First find the common acceleration, then isolate the 3 kg block.
Review the solution
  1. Principle

    The combined system has a=4 m/s².

  2. Representation

    Now isolate the 3 kg block.

  3. Calculation

    Its only horizontal force is contact C.

  4. Interpretation

    C=ma=3(4)=12 N.

Dynamics

Exercise to explore

Acceleration from a 2D resultant

Type
Numerical
Difficulty
2/5
Time
5 min

The net force on a 4 kg particle is ΣF=(12,-8) N. Find a_x and a_y.

Request a hint
  • Divide each force component by the mass.
Review the solution
  1. Principle

    Each component obeys ΣF_i=ma_i.

  2. Representation

    a_x=12/4=3 m/s².

  3. Calculation

    a_y=-8/4=-2 m/s².

  4. Interpretation

    Each acceleration component follows its corresponding net-force component.

Dynamics

Exercise to explore

Constraint of an ideal rope

Type
Conceptual
Difficulty
2/5
Time
5 min

Two masses are connected by one inextensible rope over an ideal fixed pulley. Which relation holds between the acceleration magnitudes of the rope ends?

Two hanging masses connected over a pulley show upward tension and downward weight on each mass.m₁m₂Tm₁gTm₂g

The ideal rope transmits the same T magnitude on both sides and constrains equal acceleration magnitudes; the weights belong to different systems.

Request a hint
  • Use the fixed total length of the rope.
Review the solution
  1. Principle

    Total rope length is constant.

  2. Representation

    When one side shortens by a distance, the other lengthens by the same distance.

  3. Calculation

    Differentiating twice gives equal acceleration magnitudes.

Dynamics

Exercise to explore

A result that forces a model change

Type
Conceptual
Difficulty
3/5
Time
5 min

A contact problem gives N=-15 N for a surface that can only push. What is the correct interpretation?

Request a hint
  • An ordinary surface cannot pull through a normal force.
Review the solution
  1. Principle

    An ordinary surface can push but cannot pull through N.

  2. Representation

    N<0 means the assumed constraint is no longer compatible.

Normal

Exercise to explore

Pulling with a vertical component

Type
Numerical
Difficulty
2/5
Time
5 min

A 10 kg block remains on a horizontal floor. It is pulled with a 40 N force at 30° above horizontal. There is no vertical acceleration. Find N. Use g=9.8 m/s².

Request a hint
  • Balance all vertical forces.
Review the solution
  1. Principle

    F_y=40 sin30°=20 N upward.

  2. Representation

    N+20-98=0.

  3. Calculation

    N=78 N.

Normal

Exercise to explore

Normal force on an incline

Type
Numerical
Difficulty
2/5
Time
5 min

An 8 kg block is on a 25° incline. There is no acceleration perpendicular to the plane and no other force has a perpendicular component. Find N. Use g=9.8 m/s².

A block on a ramp has vertical weight and perpendicular normal force; a dashed guide marks the perpendicular weight component.blockNmgguide: mg cosθ

Normal force is perpendicular to the plane. The mg cosθ guide is a weight component, not an additional force.

Request a hint
  • Balance forces perpendicular to the plane.
Review the solution
  1. Principle

    There is no perpendicular acceleration.

  2. Representation

    N=mg cos25°.

  3. Calculation

    N≈71.05 N.

Normal

Exercise to explore

Elevator accelerating upward

Type
Numerical
Difficulty
2/5
Time
5 min

A 60 kg person stands on a scale in an elevator accelerating upward at 1.5 m/s². Find N. Use g=9.8 m/s².

A person on a scale has upward normal force and downward weight; a separate elevator-acceleration marker appears alongside.elevatorpersonNmgelevator a

N and mg are forces on the person. The elevator-acceleration marker is separate because acceleration is not a force.

Request a hint
  • Take upward as positive.
Review the solution
  1. Principle

    Take +y upward.

  2. Representation

    N-mg=ma.

  3. Calculation

    N=m(g+a)=60(11.3)=678 N.

Normal

Exercise to explore

Apparent weightlessness

Type
Conceptual
Difficulty
2/5
Time
5 min

A person and elevator are in ideal free fall. What is the floor's normal force on the person while there is no relative support?

A person on a scale has upward normal force and downward weight; a separate elevator-acceleration marker appears alongside.elevatorpersonNmgelevator a

N and mg are forces on the person. The elevator-acceleration marker is separate because acceleration is not a force.

Request a hint
  • Apply the second law with acceleration -g.
Review the solution
  1. Principle

    With +y upward, N-mg=m(-g).

  2. Representation

    Therefore N=0.

  3. Calculation

    Gravity still acts.

Tension

Exercise to explore

Mass accelerated by a vertical rope

Type
Numerical
Difficulty
2/5
Time
5 min

A 4 kg mass accelerates upward at 2 m/s² while supported by a rope. Find T. Use g=9.8 m/s².

Request a hint
  • Take upward as positive.
Review the solution
  1. Principle

    Take upward as positive.

  2. Representation

    T-mg=ma.

  3. Calculation

    T=4(9.8+2)=47.2 N.

Tension

Exercise to explore

Acceleration in an Atwood machine

Type
Numerical
Difficulty
3/5
Time
5 min

m1=2 kg and m2=3 kg are connected by an ideal rope over an ideal pulley. Find the acceleration magnitude. Use g=9.8 m/s².

Two hanging masses connected over a pulley show upward tension and downward weight on each mass.m₁m₂Tm₁gTm₂g

The ideal rope transmits the same T magnitude on both sides and constrains equal acceleration magnitudes; the weights belong to different systems.

Request a hint
  • Write one second-law equation for each mass.
Review the solution
  1. Principle

    Write one second-law equation for each mass.

  2. Representation

    Tension cancels when the equations are added.

  3. Calculation

    a=(m2-m1)g/(m1+m2)=(1)(9.8)/5=1.96 m/s².

Tension

Exercise to explore

Tension in an Atwood machine

Type
Numerical
Difficulty
3/5
Time
5 min

For m1=2 kg, m2=3 kg and a=1.96 m/s² in the ideal Atwood machine, find T. Use g=9.8 m/s².

Two hanging masses connected over a pulley show upward tension and downward weight on each mass.m₁m₂Tm₁gTm₂g

The ideal rope transmits the same T magnitude on both sides and constrains equal acceleration magnitudes; the weights belong to different systems.

Request a hint
  • Isolate m1, which accelerates upward.
Review the solution
  1. Principle

    Isolate m1, which accelerates upward.

  2. Representation

    T-m1g=m1a.

  3. Calculation

    T=2(9.8+1.96)=23.52 N.

Tension

Exercise to explore

Nearly horizontal cables

Type
Conceptual
Difficulty
3/5
Time
5 min

The same load is supported symmetrically by two cables. What happens to each cable's tension as its angle above the horizontal decreases toward 0°?

A central lamp has two symmetric angled tensions and its downward weight.left Tright Tweight

Tensions follow the cables. Their horizontal components cancel and their vertical components support the weight, without revealing T numerically.

Request a hint
  • Use vertical equilibrium.
Review the solution
  1. Principle

    Vertical equilibrium requires 2T sinθ=mg.

  2. Representation

    T=mg/(2sinθ).

  3. Calculation

    As sinθ→0, T→∞ in the ideal model.

Friction

Exercise to explore

Required static friction

Type
Numerical
Difficulty
1/5
Time
5 min

A 10 kg block is at rest on a horizontal floor with μ_s=0.50. A 20 N horizontal force is applied and the block does not move. What is |f_s|? Use g=9.8 m/s².

Static friction adjusts up to f_s,max

Required static friction versus applied force, ending at the static maximum.
  • f_s=F_app up to f_s,max

The f_s=F_app branch ends at the threshold. Beyond it, the graph no longer represents a static state.

Request a hint
  • First calculate the friction required for equilibrium.
Review the solution
  1. Principle

    N=98 N.

  2. Representation

    f_s,max=μ_sN=49 N.

  3. Calculation

    Equilibrium requires only 20 N, so f_s=20 N.

Friction

Exercise to explore

Sliding threshold

Type
Numerical
Difficulty
2/5
Time
5 min

A 12 kg block rests on a horizontal floor with μ_s=0.40. What maximum horizontal force can static friction balance before sliding? Use g=9.8 m/s².

Request a hint
  • Use f_s,max=μ_sN.
Review the solution
  1. Principle

    N=117.6 N.

  2. Representation

    f_s,max=μ_sN.

  3. Calculation

    f_s,max=0.40(117.6)=47.04 N.

Friction

Exercise to explore

Sliding block with kinetic friction

Type
Numerical
Difficulty
2/5
Time
5 min

A 5 kg block slides in +x on a horizontal floor. μ_k=0.20 and a 20 N horizontal force acts in +x. Find a_x. Use g=9.8 m/s².

Request a hint
  • Find kinetic friction before the horizontal resultant.
Review the solution
  1. Principle

    N=49 N and f_k=9.8 N toward -x.

  2. Representation

    ΣF_x=20-9.8=10.2 N.

  3. Calculation

    a_x=10.2/5=2.04 m/s².

Friction

Exercise to explore

Friction while walking

Type
Conceptual
Difficulty
2/5
Time
5 min

While walking forward without slipping, the foot pushes the ground backward. In which direction does static friction from the ground act on the foot during that phase?

Request a hint
  • Consider the foot's tendency to slip relative to the ground.
Review the solution
  1. Principle

    The foot tends to slip backward relative to the ground.

  2. Representation

    Static friction on the foot points forward.

Friction

Exercise to explore

Can it remain at rest?

Type
Conceptual
Difficulty
3/5
Time
5 min

A block on an incline would require 30 N of static friction for equilibrium, but μ_sN=24 N. What follows from the model?

Request a hint
  • Compare required friction with the available maximum.
Review the solution
  1. Principle

    Equilibrium would require 30 N.

  2. Representation

    Static friction can reach only 24 N.

  3. Calculation

    Required friction exceeds the available maximum, so slipping begins.

Friction

Exercise to explore

Pulling upward reduces friction

Type
Numerical
Difficulty
3/5
Time
5 min

A 10 kg block slides on a horizontal floor with μ_k=0.25. It is pulled with 50 N at 37° above +x. Use sin37°=0.60, cos37°=0.80 and g=9.8 m/s². Find the kinetic-friction magnitude.

Request a hint
  • The upward component reduces the normal force.
Review the solution
  1. Principle

    F_y=50(0.60)=30 N upward.

  2. Representation

    N=98-30=68 N.

  3. Calculation

    f_k=0.25(68)=17 N.

Drag

Exercise to explore

Velocity relative to air

Type
Conceptual
Difficulty
2/5
Time
5 min

A cyclist moves east at 12 m/s relative to the ground while wind blows east at 5 m/s. What cyclist-air relative speed belongs in a simple longitudinal drag model?

Request a hint
  • Subtract the air velocity from the cyclist velocity.
Review the solution
  1. Principle

    The model uses velocity relative to air.

  2. Representation

    v_cyclist/air=12-5=7 m/s.

Drag

Exercise to explore

Linear drag

Type
Numerical
Difficulty
2/5
Time
5 min

In a linear regime, F_D=-b v_rel. For b=3.0 kg/s and v_rel=4.0 m/s in +x, find F_D,x.

Request a hint
  • The minus sign makes drag oppose relative velocity.
Review the solution
  1. Principle

    The negative sign opposes relative velocity.

  2. Representation

    F_D,x=-(3.0)(4.0)=-12 N.

Drag

Exercise to explore

Terminal speed with linear drag

Type
Numerical
Difficulty
2/5
Time
5 min

A 0.50 kg body falls with linear drag b=2.0 kg/s. Ignore buoyancy and use g=9.8 m/s². Find |v_t|.

Request a hint
  • At terminal speed, net force is zero.
Review the solution
  1. Principle

    At terminal speed the resultant is zero.

  2. Representation

    mg=bv_t.

  3. Calculation

    v_t=mg/b=(0.50)(9.8)/2.0=2.45 m/s.

Drag

Exercise to explore

Terminal speed with quadratic drag

Type
Numerical
Difficulty
3/5
Time
5 min

A 0.20 kg body falls with quadratic drag magnitude c v², where c=0.050 kg/m. Ignore buoyancy and use g=9.8 m/s². Find |v_t|.

Three fall states show the same weight and increasing drag arrows until drag equals weight at terminal speed.low speedintermediateterminalmgF_DmgF_DmgF_D

As speed increases, drag grows: first it is smaller than weight and at terminal state it matches weight. Gravity never disappears.

Request a hint
  • At terminal speed, balance weight and drag.
Review the solution
  1. Principle

    At terminal speed, mg=cv_t².

  2. Representation

    v_t=sqrt(mg/c).

  3. Calculation

    v_t=sqrt(39.2)=6.26 m/s.

Circular dynamics

Exercise to explore

Required friction on a flat curve

Type
Numerical
Difficulty
2/5
Time
5 min

A 1000 kg car takes a flat curve of radius 50 m at 10 m/s. What horizontal radial force magnitude is required?

A car on a circular path has tangent velocity and inward radial friction in a separate FBD.carFBDtangent vinward f_s

Velocity is tangent and static friction points inward. No separate centripetal force is added.

Request a hint
  • Use m v²/R v^2/R toward the centre.
Review the solution
  1. Principle

    The radial direction points inward.

  2. Representation

    F_rad=m v²/R v^2/R .

  3. Calculation

    F_rad=1000(100)/50=2000 N inward.

Circular dynamics

Exercise to explore

Maximum speed before slipping

Type
Numerical
Difficulty
3/5
Time
5 min

A flat curve has R=50 m and μ_s=0.40. Find the ideal maximum speed before slipping. Use g=9.8 m/s².

A car on a circular path has tangent velocity and inward radial friction in a separate FBD.carFBDtangent vinward f_s

Velocity is tangent and static friction points inward. No separate centripetal force is added.

Request a hint
  • At the threshold, static friction reaches μ_sN.
Review the solution
  1. Principle

    Static friction provides the radial resultant.

  2. Representation

    At the threshold, μ_smg=m v²/R v^2/R .

  3. Calculation

    v_max=sqrt(μ_sgR)=sqrt(0.40·9.8·50)=14.0 m/s.

Circular dynamics

Exercise to explore

Is there an additional centripetal force?

Type
Conceptual
Difficulty
1/5
Time
5 min

In the FBD of a car turning on a flat road, friction points inward. Should an additional arrow called ‘centripetal force’ also be drawn?

A car on a circular path has tangent velocity and inward radial friction in a separate FBD.carFBDtangent vinward f_s

Velocity is tangent and static friction points inward. No separate centripetal force is added.

Request a hint
  • Identify the actual interactions on the car.
Review the solution
  1. Principle

    ‘Centripetal’ describes the resultant's direction.

  2. Representation

    Friction is already the inward physical interaction.

  3. Calculation

    No independent force is added.

Circular dynamics

Exercise to explore

Frictionless banking angle

Type
Numerical
Difficulty
3/5
Time
5 min

A curve of radius 80 m is designed for 20 m/s without friction. Find tanθ and θ. Use g=9.8 m/s².

A banked road section shows angled normal force, vertical weight, and guides for normal-force components.vehicleNmg

Normal force is perpendicular to the road; its horizontal component points inward and its vertical component balances weight. The dashed guides separate those two components of the normal force.

Request a hint
  • Use tanθ=v²/(Rg).
Review the solution
  1. Principle

    Without friction, components of N support and accelerate the car radially.

  2. Representation

    tanθ=v²/(Rg).

  3. Calculation

    tanθ=400/784=0.5102.

  4. Interpretation

    θ=atan(0.5102)≈27.0°.

Circular dynamics

Exercise to explore

Normal force at the bottom of a circular path

Type
Numerical
Difficulty
3/5
Time
5 min

A 2 kg object passes the bottom of a circular path of radius 4 m at 6 m/s. Contact provides an upward normal force. Find N. Use g=9.8 m/s².

Top and bottom points show the local inward direction and normal and gravitational forces.inwardmg and contactinward; Nmgtopbottom

The inward direction changes: at the bottom it is upward and at the top downward. Radial equations are built locally without energy.

Request a hint
  • At the bottom, inward is upward.
Review the solution
  1. Principle

    At the bottom, inward is upward.

  2. Representation

    N-mg=m v²/R v^2/R =18 N.

  3. Calculation

    N=18+19.6=37.6 N.

Circular dynamics

Exercise to explore

Constant speed does not mean zero acceleration

Type
Conceptual
Difficulty
2/5
Time
5 min

A particle moves in a circle at constant speed. Which statement is correct?

Request a hint
  • Ask whether the velocity vector changes direction.
Review the solution
  1. Principle

    Speed does not change, but velocity direction does.

  2. Representation

    That change requires radial inward acceleration.

Fundamental forces

Exercise to explore

The four interactions

Type
Conceptual
Difficulty
1/5
Time
5 min

Which list contains only fundamental interactions?

Request a hint
  • Separate fundamental interactions from effective forces.
Review the solution
  1. Principle

    The fundamental interactions are gravitational, electromagnetic, strong, and weak.

  2. Representation

    Normal force, friction, and tension are effective forces.

Fundamental forces

Exercise to explore

Where does the normal force come from?

Type
Conceptual
Difficulty
2/5
Time
5 min

Microscopically, which fundamental interaction is mainly responsible for the contact response modelled as the normal force between solids?

Four fundamental interactions form a two-row block; below it, three everyday forces connect to the electromagnetic interaction as their effective description.gravitystrongelectromagneticweaknormalfrictiontensioneffective macroscopic forces

The upper block contains the four fundamental interactions. Normal force, friction, and tension appear below as effective macroscopic descriptions, mainly electromagnetic in origin.

Request a hint
  • Consider electron clouds in nearby matter.
Review the solution
  1. Principle

    Normal force is an effective contact force.

  2. Representation

    Responses of electron clouds and atomic structures are mainly electromagnetic.

Fundamental forces

Exercise to explore

Fundamental does not mean ‘more useful’

Type
Conceptual
Difficulty
3/5
Time
5 min

Which statement best describes an effective force such as friction?

Request a hint
  • Effective models may be useful within a stated regime.
Review the solution
  1. Principle

    Effective forces describe the relevant scale within a domain of validity.

  2. Representation

    They need not be fundamental to be physically useful.