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Kinetic and Static Friction AP  Physics 1 FRQ

Kinetic and Static Friction AP  Physics 1 FRQ – Exam Style Questions etc.

Kinetic and Static Friction AP  Physics 1 FRQ

Unit: 2. Force and Translational  Dynamics

Weightage : 10-15%

AP Physics 1 Exam Style Questions – All Topics

Exam Style Practice Questions, Kinetic and Static Friction AP  Physics 1 FRQ

Question

 A small box of mass m is placed on an inclined plane with an angle of incline of θ. There is a coefficient of kinetic friction \(\mu _{k}\) between the inclined plane and the small box. The small box is attached to a much heavier box of mass 3m by a pulley system shown below.

(a) Draw free body diagrams of both masses, including all of the forces acting on each.
(b) Assuming a frictionless, massless pulley, determine the acceleration of the blocks once they are released from rest in terms of \(\mu _{k}\), g, and θ.
(c) If \(\mu _{k}\) = 0.3 and θ = 45°, what distance is traveled by the blocks 3 s after being released from rest?

▶️Answer/Explanation

Ans:

(a) The forces acting on the small m mass are \(F _{T}\) (the tension in the string connecting it to the 3m block), \(F _{w}\) (the weight of the block), \(F _{N}\) (the normal force exerted by the inclined plane), and \(F _{f}\) (the force of kinetic friction).

 

The forces acting on the larger 3m mass are \(F _{T}\) (the tension in the string connecting it to the m block) and \(F _{w}\) (the weight of the block).

(b) Newton’s Second Law applied to the 3m mass yields

\(F_{w2}-F_{T}=3ma\Rightarrow 3mg-F_{T}=3ma\)

Newton’s Second Law applied to the m yields

\(F_{T}-F_{f}-F_{N}=ma\Rightarrow F_{T}-\mu _{k}mg\sin \theta -mg\cos \theta\)

Adding these two equations together, you find that

\(3mg-\mu _{k} mg\sin \theta -mg\cos \theta =ma+3ma\)

\(3mg-\mu _{k} mg\sin \theta -mg\cos \theta =4ma\)

\(a=\frac{3g-\mu _{k}g\sin \theta -\cos \theta }{4}\)

(c) Plugging in g = 10 m/\(s^{2}\), θ = 45°, and \(\mu _{k}\) = 1.0, the acceleration can be calculated as follows:

Apply Big Five #3: 

\( d=v_{0}t+\frac{1}{2}at^{2}\Rightarrow d=0+\frac{1}{2}(5.2m/s)^{2}=23.4m\)

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