Kinetic and Static Friction AP Physics C Mechanics FRQ – Exam Style Questions etc.
Kinetic and Static Friction AP Physics C Mechanics FRQ
Unit 2: Force and Translational Dynamics
Weightage : 20-15%
Question
A block of mass m is pulled across a rough horizontal table by a string connected to a motor that is attached to the floor. The string passes over a pulley with negligible friction that is vertically aligned with the left edge of the table as shown. The string and pulley both have negligible mass. The pulley is at height H above the table. The motor exerts a constant force of tension \(F_{T}\) on the string, and the block remains in contact with the table at all times as the block slides across the table from x = L to x = 0. The coefficient of kinetic friction between the table and the block is \(\mu _{k}\). Express all algebraic answers in terms of m, H, \(F _{T}\) , x, \(\mu _{k}\), L, and physical constants as appropriate.
(a) On the dot below that represents the block, draw and label the forces (not components) that act on the block when the block is at x = L. Each force must be represented by a distinct arrow starting on, and pointing away from, the dot.
(b) Derive an expression for the angle \(\Theta \) that the string makes with the horizontal as a function of x.
(c)
i. Derive an expression for the normal force \(F_{N}\) exerted on the block by the table as a function of the block’s position x.
ii. Derive an expression for the magnitude of the net horizontal force \(F_{net}\) exerted on the block as a function of the position x.
(d) Write, but do not solve, an integral expression that could be used to solve for the work W done by the string on the block as the block moves from x = L to x = 0.
(e) Does the string do more, less, or the same amount of work on the block as the block moves from x = L to \(x = \frac{L}{2}\) compared to when the block moves from x = L to \(x = \frac{L}{2}\) to x = 0 ?
______ More work when the block moves from x = L to \(x = \frac{L}{2}\)
______ Less work when the block moves from x = L to \(x = \frac{L}{2}\)
______The same amount of work when the block moves from x = L to \(x = \frac{L}{2}\)
Justify your answer
▶️Answer/Explanation
1(a) Example Response
- Scoring Note: Examples of appropriate labels for the force due to gravity include: \(F_{G}\), \(F_{g}\), \(F_{grav}\), W , mg , Mg , “grav force”, “F Earth on block”, “F on block by Earth”, \(F_{Earth\\\ on\\\ Block}\), \(F_{E\\\ ,\\\ Block}\). The labels G and g are not appropriate labels for the force due to gravity. \(F_{n}\), \(F_{N}\), N , “normal force”, “ground force”, or similar labels may be used for the normal force. \(F_{string}\), \(F_{s}\), \(F_{T}\), \(F_{Tension}\), T, “string force,” “tension force,” or similar labels may be used for the tension force exerted by the string.
- A response with extraneous forces or vectors can earn a maximum of two points.
1(b) Example Responses
1(c)(i) Example Responses
For using Newton’s second law to sum the forces in the vertical direction and write an equation that is consistent with part (a)
\(\sum F_{y} = ma\)
\(F_{net, y} = F_{N} + F_{T,y} – F_{g} = 0\)
\(F_{N} = F_{g} – F_{T, y}\)
For correctly substituting the vertical component of the tension in terms of the given variables consistent with part (b)
\(F_{N} = mg -F_{T} sin\Theta\)
\(F_{N} = mg -F_{T}\left ( \frac{H}{\sqrt{H^{2} + x^{2}}} \right )\)
1(c)(ii) Example Response
For using Newton’s second law to sum the forces in the horizontal direction and write an equation that is consistent with part (a)
\(F_{net, x} = F_{T, x} – F_{f}\)
For correctly substituting the horizontal component of the force of tension in terms of the given variables consistent with part (b)
\( F_{net, x} = F_{T}cos\Theta – F_{f}\)
\(F_{net, x} = F_{T}(\frac{(x)}{\sqrt{H^{2}+x^{2}}}) – F_{f}\)
For correctly substituting the expression for the normal force from part (c)(i) into the expression for the force of friction
\(F_{net, x} = F_{T}(\frac{(x)}{\sqrt{H^{2}+x^{2}}}) – \mu _{k}F_{N}\)
\(F_{net, x} = F_{T}(\frac{(x)}{\sqrt{H^{2}+x^{2}}}) – \mu _{k}(mg – F_{r(\frac{H}{\sqrt{H^{2}+x^{2}}})})\)
1(d)Example Response
For any indication that the work done on the block by the string is due only to the horizontal component of the tension in the string
\( W = \int F_{T, x}dx\)
For using the horizontal component of the force of tension consistent with part (c)
For indicating that work is the integral of the force with respect to x, including limits or a constant of integration
\( W = \int_{x=L}^{x = 0} – F_{T}(\frac{x}{\sqrt{H^{2}+x^{2}}})dx\)
1(e)Example Response
\(F_{T}\) stays the same for both halves, the displacement is the same in both halves, but from x = L to x = L /2 the angle is smaller, resulting in a larger component of the tension force that aligns with the displacement
Question
A 300-kg box rests on a platform attached to a forklift, shown above. Starting from rest at time = 0, the box is lowered with a downward acceleration of 1.5 m/s2
a. Determine the upward force exerted by the horizontal platform on the box as it is lowered.
At time t = 0, the forklift also begins to move forward with an acceleration of 2 m/s2 while lowering the box as described above. The box does not slip or tip over.
b. Determine the frictional force on the box.
c. Given that the box does not slip, determine the minimum possible coefficient of friction between the box and the platform.
d. Determine an equation for the path of the box that expresses y as a function of x (and not of t), assuming that, at time t = 0, the box has a horizontal position x = 0 and a vertical position y = 2 m above the ground, with zero velocity.
e. On the axes below sketch the path taken by the box
Answer/Explanation
Ans:
a. ΣF = ma; using downward as the positive direction, mg – N = may gives N = m(g – ay) = 2490 N
b. Friction is the only horizontal force exerted; ΣF = f = max = 600 N
c. At the minimum coefficient of friction, static friction will be at its maximum value f = µN, giving µ = f/N = (600 N)/(2490 N) = 0.24
d. y = y0 + v0yt + ½ ayt2 = 2 m + ½ (–1.5 m/s2)t2 and x = x0 + v0xt + ½ axt2 = ½ (2 m/s2)t2, solving for t2 in the x equation gives t2 = x. Substituting into the y equation gives y as a function of x: y = 2 – 0.75x
e.