Statics and Dynamics Quiz: Computing Work
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Computing WorkQuestion 1 of 1

A 10 kg crate is pushed up a 30°30° incline by a horizontal force FhF_h. The kinetic friction coefficient is μk=0.25\mu_k = 0.25. The crate moves a distance of d=6 md = 6 \text{ m} along the incline surface. Take g=9.81 m/s2g = 9.81 \text{ m/s}^2. The horizontal force magnitude is Fh=80 NF_h = 80 \text{ N}.

What is the work done by the friction force on the crate over the 6 m path along the incline?

Wf=−μk(mgcos⁡30°+Fhsin⁡30°)⋅d=−0.25(84.96+40)(6)≈−187.4 JW_f = -\mu_k(mg\cos30° + F_h\sin30°)\cdot d = -0.25(84.96 + 40)(6) \approx -187.4 \text{ J}, because the normal force on the incline is increased by the component of the horizontal push perpendicular to the incline surface.
Wf=−μk(mgcos⁡30°)⋅d=−0.25(84.96)(6)≈−127.4 JW_f = -\mu_k(mg\cos30°)\cdot d = -0.25(84.96)(6) \approx -127.4 \text{ J}, because only the weight component perpendicular to the incline contributes to the normal force, and the horizontal force is parallel to the incline.
Wf=−μk(mg)⋅d=−0.25(98.1)(6)=−147.2 JW_f = -\mu_k(mg)\cdot d = -0.25(98.1)(6) = -147.2 \text{ J}, because friction depends on the full weight of the crate and acts over the full path length along the incline.
Wf=−μk(mgcos⁡30°−Fhsin⁡30°)⋅d=−0.25(84.96−40)(6)≈−67.4 JW_f = -\mu_k(mg\cos30° - F_h\sin30°)\cdot d = -0.25(84.96 - 40)(6) \approx -67.4 \text{ J}, because the horizontal push has a component along the incline surface that reduces the effective normal force by its perpendicular-to-surface projection.
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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Computing Work

Practice Computing Work in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

A 10 kg crate is pushed up a 30°30° incline by a horizontal force FhF_h. The kinetic friction coefficient is μk=0.25\mu_k = 0.25. The crate moves a distance of d=6 md = 6 \text{ m} along the incline surface. Take g=9.81 m/s2g = 9.81 \text{ m/s}^2. The horizontal force magnitude is Fh=80 NF_h = 80 \text{ N}.

What is the work done by the friction force on the crate over the 6 m path along the incline?

  1. Wf=−μk(mgcos⁡30°+Fhsin⁡30°)⋅d=−0.25(84.96+40)(6)≈−187.4 JW_f = -\mu_k(mg\cos30° + F_h\sin30°)\cdot d = -0.25(84.96 + 40)(6) \approx -187.4 \text{ J}, because the normal force on the incline is increased by the component of the horizontal push perpendicular to the incline surface. (correct answer)
  2. Wf=−μk(mgcos⁡30°)⋅d=−0.25(84.96)(6)≈−127.4 JW_f = -\mu_k(mg\cos30°)\cdot d = -0.25(84.96)(6) \approx -127.4 \text{ J}, because only the weight component perpendicular to the incline contributes to the normal force, and the horizontal force is parallel to the incline.
  3. Wf=−μk(mg)⋅d=−0.25(98.1)(6)=−147.2 JW_f = -\mu_k(mg)\cdot d = -0.25(98.1)(6) = -147.2 \text{ J}, because friction depends on the full weight of the crate and acts over the full path length along the incline.
  4. Wf=−μk(mgcos⁡30°−Fhsin⁡30°)⋅d=−0.25(84.96−40)(6)≈−67.4 JW_f = -\mu_k(mg\cos30° - F_h\sin30°)\cdot d = -0.25(84.96 - 40)(6) \approx -67.4 \text{ J}, because the horizontal push has a component along the incline surface that reduces the effective normal force by its perpendicular-to-surface projection.
Explanation: Whenever a force is applied at an angle to an inclined surface, your first job is to resolve every force into components perpendicular and parallel to the incline — because the normal force (and therefore friction) depends only on the net perpendicular equilibrium. Here, the horizontal force Fh=80 NF_h = 80\text{ N} has two incline-frame components: Fhcos⁡30°F_h\cos30° along the incline (helping push the crate up) and Fhsin⁡30°F_h\sin30° perpendicular into the incline surface (pressing the crate harder against the incline). Setting up the perpendicular equilibrium gives: N=mgcos⁡30°+Fhsin⁡30°=(10)(9.81)(0.866)+(80)(0.5)=84.96+40=124.96 NN = mg\cos30° + F_h\sin30° = (10)(9.81)(0.866) + (80)(0.5) = 84.96 + 40 = 124.96\text{ N} The kinetic friction force is then fk=μkN=0.25(124.96)=31.24 Nf_k = \mu_k N = 0.25(124.96) = 31.24\text{ N}, opposing motion (directed down the incline). Over d=6 md = 6\text{ m}, the work done by friction is: Wf=−fk⋅d=−0.25(124.96)(6)≈−187.4 JW_f = -f_k \cdot d = -0.25(124.96)(6) \approx -187.4\text{ J} This confirms A is correct. Now for the traps: B ignores the horizontal force's contribution to the normal force entirely — it treats FhF_h as if it were parallel to the incline, which it isn't. C uses the full weight mgmg instead of the perpendicular weight component mgcos⁡30°mg\cos30°, confusing the incline geometry. D subtracts Fhsin⁡30°F_h\sin30° from the normal force — this would only make sense if FhF_h were pulling the crate away from the surface, but here the horizontal push drives the crate into the incline, so that component adds to NN, not subtracts. Your study tip: always draw a free-body diagram in incline coordinates first. Ask yourself — does each force's perpendicular component push the object into the surface or away from it? That sign determines whether NN increases or decreases, and friction follows from there.