Practice · Unit 4 · Aula Física

Work

A constant 42 N horizontal force moves a box 3.5 m in the same direction. Calculate the work by that force.

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Use θ=0°.

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Principle
W=FΔx cosθ.
Representation
F and Δx point in the same direction.
Calculation
W=(42)(3.5)cos0°=147 J.
Interpretation
The work is positive.

Work

A 70 N force acts on a sled that moves 4.0 m horizontally. The force makes 60° with displacement. Calculate its work.

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Only the parallel component contributes.

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Principle
W=FΔr cosθ.
Representation
θ=60° between F and Δr.
Calculation
W=(70)(4.0)cos60°=140 J.
Interpretation
The perpendicular component does no work.

Work

An 18 N kinetic-friction force opposes a box's motion for 6.2 m. Calculate the work by friction.

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The angle is 180°.

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Principle
W=f_kd cos180°.
Representation
Friction and displacement are opposite.
Calculation
W=(18)(6.2)(-1)=-111.6 J.
Interpretation
Friction removes mechanical energy from the box.

Work

A block moves horizontally while a 25 N force is exactly vertical. What work does that force do?

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Evaluate the dot product.

Review the solution

Principle
F and Δr are perpendicular.
Representation
cos90°=0.
Calculation
F·Δr=0.

Work

A box moves 4.0 m. A 50 N force acts 37° above horizontal and friction is 12 N opposite motion. Weight and normal do no work. Calculate net work.

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Calculate applied-force work and friction work separately.

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Principle
W_net=W_ap+W_f.
Representation
W_ap=(50)(4.0)cos37°≈159.7 J and W_f=-48 J.
Calculation
W_net≈111.7 J.
Interpretation
Positive net work increases K.

Kinetic energy

Calculate the kinetic energy of a 1.8 kg particle moving at 6.0 m/s.

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Square the speed.

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Principle
$K=\frac12mv^2$ .
Representation
m=1.8 kg and v=6.0 m/s.
Calculation
K=(1/2)(1.8)(36)=32.4 J.
Interpretation
K is nonnegative.

Kinetic energy

A 2.0 kg particle increases its speed from 3.0 m/s to 7.0 m/s. Calculate ΔK.

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Subtract K_i from K_f.

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Principle
ΔK=(1/2)m(v_f²-v_i²).
Representation
Substitute both speeds.
Calculation
ΔK=(1/2)(2)(49-9)=40 J.
Interpretation
The positive change corresponds to greater speed.

Kinetic energy

A 3.0 kg cart starts at 2.0 m/s and receives 96 J of net work. Calculate its final speed.

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Find K_f first.

Review the solution

Principle
W_net=K_f-K_i.
Representation
K_i=6 J and K_f=102 J.
Calculation
v_f=sqrt(2K_f/m)=sqrt68≈8.25 m/s.
Interpretation
Positive work increases speed.

Kinetic energy

A 60 kg body moves at 8.0 m/s. A constant 240 N net force opposite motion acts until it stops. Calculate stopping distance.

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Net work is -Fd.

Review the solution

Principle
K_i=(1/2)(60)(8²)=1920 J.
Representation
K_f=0, so W_net=-1920 J.
Calculation
-240d=-1920 gives d=8.0 m.
Interpretation
Distance is positive although work is negative.

Kinetic energy

Net work on a particle is zero between two points. What is necessarily true for constant mass?

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Use ΔK=0.

Review the solution

Principle
$W_{net}=\Delta K$ =0.
Representation
K depends on v².
Calculation
Final speed equals initial speed, although direction may change.

Variable force

A force component is F_x=4x N with x in metres. Calculate work from x=0 to x=3.0 m.

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Integrate 4x with respect to x.

Review the solution

Principle
W=∫_0^3 4x dx.
Representation
An antiderivative is 2x².
Calculation
W=[2x²]_0^3=18 J.
Interpretation
The area is positive.

Variable force

Force varies as F_x=12-2x N. Calculate work from x=1.0 m to x=5.0 m.

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Evaluate the antiderivative at both limits.

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Principle
W=∫_1^5(12-2x)dx.
Representation
The antiderivative is 12x-x².
Calculation
W=(60-25)-(12-1)=24 J.
Interpretation
The limits preserve displacement direction.

Variable force

A force is +6 N from x=0 to 2 m and then -2 N from x=2 to 5 m. Calculate total work.

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Keep the sign of each rectangle.

Review the solution

Principle
Work is algebraic area.
Representation
W_1=(6)(2)=12 J and W_2=(-2)(3)=-6 J.
Calculation
W=12-6=6 J.
Interpretation
Absolute areas are not added.

Variable force

An ideal k=200 N/m spring changes from x_i=0.10 m to x_f=0.25 m. Calculate work by the spring.

Request a hint

Use the difference of squares with a minus sign.

Review the solution

Principle
W_s=-(1/2)k(x_f²-x_i²).
Representation
Use deformations from natural length.
Calculation
W_s=-100(0.0625-0.0100)=-5.25 J.
Interpretation
The spring opposes increased deformation.

Variable force

On an F_x versus x graph, force forms a positive triangle from x=0 to 4 m with height 10 N at x=2 m. Calculate work.

Adding positive and negative areas

positive segment
negative segment

The positive segment contributes +12 J and the negative segment -6 J; the illustrated total is +6 J.

Request a hint

Use the area of a triangle.

Review the solution

Principle
Work is signed area under the graph.
Representation
The base is 4 m and height 10 N.
Calculation
W=(1/2)(4)(10)=20 J.
Interpretation
The region is above the axis and contributes positively.

Power

A device does 3600 J of work in 18 s. Calculate average power.

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Divide work by time.

Review the solution

Principle
P_avg=W/Δt.
Representation
W=3600 J and Δt=18 s.
Calculation
P_avg=3600/18=200 W.
Interpretation
A watt is a joule per second.

Power

An 800 kg elevator rises at constant 1.5 m/s. Ignore losses and use g=9.8 m/s². Calculate motor mechanical power.

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At constant speed the lifting force balances mg.

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Principle
At constant speed F=mg.
Representation
P=Fv=mgv.
Calculation
P=(800)(9.8)(1.5)=11760 W.
Interpretation
This is 11.76 kW.

Power

A 120 N force acts on a cart moving at 2.5 m/s. The force-velocity angle is 30°. Calculate instantaneous power.

Request a hint

Include cos30°.

Review the solution

Principle
P=Fv cosθ.
Representation
F=120 N, v=2.5 m/s, θ=30°.
Calculation
P≈259.8 W.
Interpretation
Only the parallel component delivers energy.

Power

Two machines do 900 J. A takes 6 s and B takes 15 s. Which statement is correct?

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Calculate W/Δt for each machine.

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Principle
Use P_avg=W/Δt.
Representation
P_A=900/6=150 W.
Calculation
P_B=900/15=60 W.
Interpretation
A transfers the same energy faster.

Power

A device operates at 1.2 kW for 45 min. Calculate transferred energy in joules and verify that it equals 0.90 kWh.

Request a hint

Convert 45 min to 2700 s.

Review the solution

Principle
E=PΔt.
Representation
1.2 kW=1200 W and 45 min=2700 s.
Calculation
E=(1200)(2700)=3.24×10^6 J.
Interpretation
This is also 0.90 kWh; kWh is energy.

Potential energy

A 3.2 kg mass rises 5.5 m. Calculate ΔU_g using g=9.8 m/s².

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Use $\Delta U_g=mg\Delta y$ .

Review the solution

Principle
$\Delta U_g=mg\Delta y$ .
Representation
The rise makes Δy positive.
Calculation
ΔU_g=(3.2)(9.8)(5.5)=172.48 J.
Interpretation
The result does not depend on the chosen zero.

Potential energy

Two students choose different U_g=0 levels and analyse the same motion from A to B. What must be equal?

Request a hint

A constant cancels from a difference.

Review the solution

Principle
Changing zero adds a constant to U.
Representation
The constant appears in both states.
Calculation
U_B-U_A does not change.

Potential energy

An ideal spring has k=300 N/m and deformation 0.080 m. Calculate U_s with U=0 at x=0.

Request a hint

Use $U_s=\frac12kx^2$ .

Review the solution

Principle
$U_s=\frac12kx^2$ .
Representation
x is measured from natural length.
Calculation
U_s=(1/2)(300)(0.080²)=0.96 J.
Interpretation
Energy is nonnegative for this reference.

Potential energy

For a k=240 N/m spring, deformation changes from x=-0.050 m to x=+0.100 m. Calculate ΔU_s.

Request a hint

Square both deformations.

Review the solution

Principle
ΔU_s=(1/2)k(x_f²-x_i²).
Representation
The sign of x vanishes when squared.
Calculation
ΔU_s=120(0.0100-0.0025)=0.90 J.
Interpretation
Final deformation has greater magnitude.

Potential energy

A 0.40 kg body descends 0.60 m while a k=100 N/m spring changes from x=0 to x=0.050 m. Calculate total ΔU using g=9.8 m/s².

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Add ΔU_g and ΔU_s with signs.

Review the solution

Principle
ΔU=ΔU_g+ΔU_s.
Representation
ΔU_g=(0.40)(9.8)(-0.60)=-2.352 J.
Calculation
ΔU_s=(1/2)(100)(0.050²)=0.125 J.
Interpretation
ΔU=-2.227 J.

Conservation

A particle starts from rest 3.2 m above a final point. Without losses, calculate final speed using g=9.8 m/s².

Request a hint

Set mgh equal to (1/2)mv².

Review the solution

Principle
$K_i+U_i=K_f+U_f$ .
Representation
Choose U_f=0 and K_i=0.
Calculation
v=sqrt(2gh)=sqrt(2·9.8·3.2)≈7.92 m/s.
Interpretation
Mass cancels.

Conservation

A 0.30 kg block starts from rest against a k=180 N/m spring compressed 0.15 m. Without losses, calculate speed at x=0.

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Conserve K+U_s.

Review the solution

Principle
Mechanical energy is conserved.
Representation
(1/2)kx²=(1/2)mv².
Calculation
v=0.15sqrt(180/0.30)≈3.67 m/s.
Interpretation
At x=0 the chosen elastic potential is zero.

Conservation

A system initially has 125 J of mechanical energy. Forces not included in U do -32 J of work. Calculate final mechanical energy.

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Add signed work to E_i.

Review the solution

Principle
ΔE_mech=W_other.
Representation
E_f-E_i=-32 J.
Calculation
E_f=125-32=93 J.
Interpretation
Mechanical energy decreases.

Conservation

A 5.0 kg block starts from rest and ends 2.0 m lower. Friction does -30 J. Calculate final speed using g=9.8 m/s².

Request a hint

Use K_f=U_i+W_f.

Review the solution

Principle
Choose U_f=0.
Representation
U_i=mgh=98 J.
Calculation
K_f=98-30=68 J.
Interpretation
v=sqrt(2·68/5)≈5.22 m/s.

Conservation

A system's mechanical energy decreases because of friction. Which description is correct?

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Track all forms of energy.

Review the solution

Principle
K+U is mechanical energy.
Representation
Friction can increase E_int.
Calculation
Total energy is conserved for an appropriate boundary.

Force and potential

Potential energy is U(x)=3x² J with x in metres. Calculate F_x at x=2.0 m.

Request a hint

Apply the minus sign after differentiating.

Review the solution

Principle
$F_x=-dU/dx$ .
Representation
dU/dx=6x.
Calculation
F_x=-6(2)=-12 N.
Interpretation
Force points toward lower U.

Force and potential

U(x)=4x-x² J. Where is equilibrium and how is it classified?

Request a hint

Find dU/dx=0 and inspect U''.

Review the solution

Principle
dU/dx=4-2x.
Representation
Slope vanishes at x=2 m.
Calculation
U''=-2<0: this is a maximum.
Interpretation
Equilibrium is unstable.

Force and potential

U(x)=(x-1)² J. Which statement is correct?

Request a hint

Locate the parabola's minimum.

Review the solution

Principle
U has zero slope at x=1 m.
Representation
There U reaches a minimum.
Calculation
A local minimum is stable equilibrium.

Force and potential

A constant conservative force F_x=+5 N acts in one dimension. Calculate ΔU from x=1 m to x=4 m.

Request a hint

Calculate W_c first.

Review the solution

Principle
W_c=FΔx.
Representation
Δx=3 m, so W_c=15 J.
Calculation
ΔU=-W_c=-15 J.
Interpretation
Potential energy decreases along the force.

Force and potential

In a region, U(x) decreases as x increases. What is the sign of F_x?

Request a hint

Reverse the slope's sign.

Review the solution

Principle
dU/dx<0.
Representation
$F_x=-dU/dx$ .
Calculation
Therefore F_x>0.

Energy diagrams

A particle moves with U(x)=2x² J and E=18 J. Find both turning points.

Request a hint

At a turning point, U=E.

Review the solution

Principle
K=0 implies U=E.
Representation
2x²=18.
Calculation
x=±3 m.
Interpretation
The turning points bound the allowed region.

Energy diagrams

A 2.0 kg particle has E=20 J and U(x)=4x² J. Calculate speed at x=1.0 m.

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First calculate $K=E-U$ .

Review the solution

Principle
U(1)=4 J.
Representation
K=20-4=16 J.
Calculation
v=sqrt(2K/m)=sqrt16=4.0 m/s.
Interpretation
The position is allowed because U<E.

Energy diagrams

A barrier has U_max=12 J. A classical particle has E=9 J with no non-conservative forces. Can it cross?

Request a hint

Use $K=E-U$ .

Review the solution

Principle
E remains fixed.
Representation
In part of the barrier U>9 J.
Calculation
There $K=E-U$ would be negative.
Interpretation
The region is inaccessible in the classical model.

Energy diagrams

U(x) has two local minima separated by a local maximum. Which classification is correct?

Request a hint

Relate minima and maxima to stability.

Review the solution

Principle
At each local extremum the slope is zero.
Representation
Minima restore small perturbations.
Calculation
The maximum amplifies perturbations.
Interpretation
The minima are stable and the maximum unstable.

Energy diagrams

For a 1D conservative particle with mechanical energy E, a position x is classically allowed when:

Request a hint

Require K≥0.

Review the solution

Principle
$K=E-U$ .
Representation
Kinetic energy cannot be negative.
Calculation
Therefore E-U≥0.
Interpretation
The condition is $U\le E$ .

Work

Kinetic energy

Variable force

Power

Potential energy

Conservation

Force and potential

Energy diagrams