Computing Work - Statics and Dynamics

1. What is the work done by the friction force on the crate over the 6 m path along the incline?
   Answer: $W_f = -\mu_k(mg\cos30° + F_h\sin30°)\cdot d = -0.25(84.96 + 40)(6) \approx -187.4 \text{ J}$, because the normal force on the incline is increased by the component of the horizontal push perpendicular to the incline surface. 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 $F_h = 80\text{ N}$ has two incline-frame components: $F_h\cos30°$ along the incline (helping push the crate up) and $F_h\sin30°$ perpendicular into the incline surface (pressing the crate harder against the incline). Setting up the perpendicular equilibrium gives: $N = mg\cos30° + F_h\sin30° = (10)(9.81)(0.866) + (80)(0.5) = 84.96 + 40 = 124.96\text{ N}$ The kinetic friction force is then $f_k = \mu_k N = 0.25(124.96) = 31.24\text{ N}$, opposing motion (directed down the incline). Over $d = 6\text{ m}$, the work done by friction is: $W_f = -f_k \cdot d = -0.25(124.96)(6) \approx -187.4\text{ J}$ This confirms A is correct. Now for the traps: B ignores the horizontal force's contribution to the normal force entirely — it treats $F_h$ as if it were parallel to the incline, which it isn't. C uses the full weight $mg$ instead of the perpendicular weight component $mg\cos30°$, confusing the incline geometry. D subtracts $F_h\sin30°$ from the normal force — this would only make sense if $F_h$ were pulling the crate away from the surface, but here the horizontal push drives the crate into the incline, so that component adds to $N$, 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 $N$ increases or decreases, and friction follows from there.

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