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AP · AP 1 · free response

AP Physics 1: Algebra-Based · Question 7

AP Physics 1: Algebra-Based · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
demanding
Marks
12
Topics
2
Answer
Complete
Block on a rough incline of angle θ leading onto a rough horizontal surfacemhθμμreleased from rest
Fig. 7.1A wedge-shaped incline stands on a horizontal floor with its sloping face rising from right to left, and the angle between the sloping face and the floor at the foot of the slope is marked θ. A block labelled m rests on the sloping face near the top and is noted as released from rest, with a dimension line to the left of the wedge marking its height h above the floor. The symbol μ is printed on the wedge below the sloping face and again on the floor beyond the foot of the slope, showing that the same coefficient of kinetic friction applies to both surfaces, which run into one another at the bottom of the incline.
free response12 marks

A block of mass m is released from rest at a height h on an incline that makes an angle θ with the horizontal. The coefficient of kinetic friction between the block and the incline is μ. The block slides down the incline and then continues onto a horizontal surface with the same coefficient of kinetic friction.

  1. (a)

    Derive Derive an expression for the speed of the block at the bottom of the incline, in terms of m, h, θ, μ and physical constants.

    4 marks
  2. (b)

    Calculate Calculate the speed of the block at the bottom of the incline for m = 2.0 kg, h = 1.5 m, θ = 30°, and μ = 0.25.

    3 marks
  3. (c)

    Determine Determine the distance the block travels along the horizontal surface before coming to rest.

    3 marks
  4. (d)

    Justify A student claims that doubling the mass of the block would double the distance found in part (c). Justify why the student is incorrect, referring to your expression.

    2 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overviewKey answer: 1 point: uses energy conservation with a friction term, ΔK = mgh − f·L 1 point: identifies the distance along the incline as L = h / sin θ and the friction force as f = μmg cos θ 1 point: friction work = μmg cos θ · (h / sin θ) = μmgh cot θ
01

(a)

4 marks

Derive Derive an expression for the speed of the block at the bottom of the incline, in terms of m, h, θ, μ and physical constants.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: uses energy conservation with a friction term, ΔK = mgh − f·L

  2. 2

    1 point: identifies the distance along the incline as L = h / sin θ and the friction force as f = μmg cos θ

  3. 3

    1 point: friction work = μmg cos θ · (h / sin θ) = μmgh cot θ

  4. 4

    1 point: v = √(2gh(1 − μ cot θ)), with m correctly cancelling

02

(b)

3 marks

Calculate Calculate the speed of the block at the bottom of the incline for m = 2.0 kg, h = 1.5 m, θ = 30°, and μ = 0.25.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: cot 30° = 1.73, so 1 − 0.25(1.73) = 0.567

  2. 2

    1 point: v = √(2 × 9.8 × 1.5 × 0.567)

  3. 3

    1 point: v = 4.1 m/s, with unit

03

(c)

3 marks

Determine Determine the distance the block travels along the horizontal surface before coming to rest.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: on the horizontal surface the only horizontal force is friction, μmg

  2. 2

    1 point: ½mv² = μmg·d, so d = v²/(2μg)

  3. 3

    1 point: d = 16.7 / (2 × 0.25 × 9.8) = 3.4 m

04

(d)

2 marks

Justify A student claims that doubling the mass of the block would double the distance found in part (c). Justify why the student is incorrect, referring to your expression.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    1 point: mass appears in both the kinetic energy and the friction force and cancels from d = v²/(2μg)

  2. 2

    1 point: v itself is also independent of m, so the distance is unchanged when the mass is doubled

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