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Physics using incline planes

The carton  lies on a plane tilted at an angle θ = 24.5° to the horizontal, with µk = 0.07.


(a) Determine the acceleration of the carton as it slides down the plane.

(b) If the carton starts from rest 8.80 m up the plane from its base, what will be the carton's speed when it reaches the bottom of the incline?

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1 Answer

(a). The acceleration of the carton is found by calculating the net force on the carton divided by its mass, ∑F/m.  There are four forces, two of which cancel each other out.

Fpar = W sin θ = mg sin θ.  [force parallel to the incline.]

Fperp = W cos θ = mg cos θ.  [force perpendicular to the incline.]

Fnor = Fperp [normal force, opposite in direction to Fperp.]

Ffric = µk * Fnor [force due to friction, opposite in direction to Fpar.]

a = ∑F/m = (Fpar - Ffric)/m = (mg sin θ - µk * mg cos θ)/m = g sin θ - µk * g cos θ = g(sin θ - µk * cos θ).

a = (9.81)(sin 24.5 - 0.07*cos 24.5) = 3.443 m/s2.

(b). Try this one considering the acceleration has been calculated in (a).

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