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In this question use $g = 9.8 \, ext{ms}^{-2}$ A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the horizontal - AQA - A-Level Maths: Pure - Question 19 - 2022 - Paper 2

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Question 19

In-this-question-use-$g-=-9.8-\,--ext{ms}^{-2}$--A-rough-wooden-ramp-is-10-metres-long-and-is-inclined-at-an-angle-of-25°-above-the-horizontal-AQA-A-Level Maths: Pure-Question 19-2022-Paper 2.png

In this question use $g = 9.8 \, ext{ms}^{-2}$ A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the horizontal. The bottom of the ram... show full transcript

Worked Solution & Example Answer:In this question use $g = 9.8 \, ext{ms}^{-2}$ A rough wooden ramp is 10 metres long and is inclined at an angle of 25° above the horizontal - AQA - A-Level Maths: Pure - Question 19 - 2022 - Paper 2

Step 1

19 (a) The tension in the rope is 230 N. The crate accelerates up the ramp at 1.2 m/s². Find the coefficient of friction between the crate and the ramp.

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Answer

To find the coefficient of friction, we first need to resolve the forces acting on the crate along the ramp and perpendicular to the ramp.

  1. Resolve Weight: The weight of the crate can be resolved into two components:

    • Parallel to the ramp:
      Wextparallel=mgextsinhetaW_{ ext{parallel}} = mg \, ext{sin} \, heta
    • Perpendicular to the ramp: Wextperpendicular=mgextcoshetaW_{ ext{perpendicular}} = mg \, ext{cos} \, heta Here, ( m = 20 , ext{kg} ), ( g = 9.8 , ext{ms}^{-2} ), and ( heta = 25° ). Therefore, Wextparallel=20imes9.8imesextsin(25°)W_{ ext{parallel}} = 20 imes 9.8 imes ext{sin}(25°) Wextperpendicular=20imes9.8imesextcos(25°)W_{ ext{perpendicular}} = 20 imes 9.8 imes ext{cos}(25°)
  2. Applying Newton's Second Law: The net force equation along the ramp can be written as: TWextparallelFextfriction=maT - W_{ ext{parallel}} - F_{ ext{friction}} = ma Where:

    • ( T = 230 , ext{N} )
    • ( F_{ ext{friction}} = , ext{μ} R = , ext{μ} mg , ext{cos}(25°) )

    Thus, substituting into the equation: 230(20imes9.8imesextsin(25°))extμ(20imes9.8×extcos(25°))=20(1.2)230 - (20 imes 9.8 imes ext{sin}(25°)) - ext{μ}(20 imes 9.8 \times ext{cos}(25°)) = 20(1.2)

  3. Solve for extμ ext{μ}: After substitution and simplification, we can find ( ext{μ} ) Rearranging to solve for ( ext{μ} ): 230196extsin(25°)20extμ(9.8extcos(25°))=24230 - 196 ext{sin}(25°) - 20 ext{μ} (9.8 ext{cos}(25°)) = 24 This leads to an expression to calculate ( ext{μ} ) yielding ( ext{μ} , ext{≈} , 0.69 ).

Step 2

19 (b) (i) The crate takes 3.8 seconds to reach the top of the ramp. Find the distance OA.

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Answer

Using the kinematic equation for uniformly accelerated motion, s=ut+12at2s = ut + \frac{1}{2} at^2 where:

  • ( u = 0 ) (initial velocity)
  • ( a = 1.2 , ext{ms}^{-2} )
  • ( t = 3.8 , ext{s} )

Substituting these values into the equation: s=0(3.8)+12(1.2)(3.82)s = 0(3.8) + \frac{1}{2}(1.2)(3.8^2) = \frac{1}{2}(1.2)(14.44) \approx 8.664 , ext{m} $$ Therefore, the distance OA is approximately ( 10 , ext{m} - 8.664 , ext{m} = 1.336 , ext{m} ).

Step 3

19 (b) (ii) Other than air resistance, state one assumption you have made about the crate in answering part (b)(i).

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Answer

One assumption made is that the crate is treated as a particle, meaning we neglect its dimensions and consider all its mass concentrated at a point. This simplifies the dynamics involved in the motion up the ramp.

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