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A particle P of mass 3 kg is projected up a line of greatest slope of a rough plane inclined at an angle of 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 6 - 2003 - Paper 1

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A particle P of mass 3 kg is projected up a line of greatest slope of a rough plane inclined at an angle of 30° to the horizontal. The coefficient of friction betwee... show full transcript

Worked Solution & Example Answer:A particle P of mass 3 kg is projected up a line of greatest slope of a rough plane inclined at an angle of 30° to the horizontal - Edexcel - A-Level Maths Mechanics - Question 6 - 2003 - Paper 1

Step 1

(a) the frictional force acting on P as it moves up the plane

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Answer

To find the frictional force acting on particle P, we need to consider the forces acting on it. The weight of P can be resolved into two components: one parallel to the incline and one perpendicular to it.

  1. Weight Component Calculation:

    The weight of P, W, is given by: W=mg=3imes9.81=29.43extNW = mg = 3 imes 9.81 = 29.43 ext{ N}

    The component of weight acting down the slope: W_{ ext{parallel}} = mg imes ext{sin}(30°) = 29.43 imes rac{1}{2} = 14.715 ext{ N}

    The normal reaction force, R, acting perpendicular to the plane is given by: R = mg imes ext{cos}(30°) = 29.43 imes rac{ ext{√3}}{2} ≈ 25.48 ext{ N}

  2. Frictional Force Calculation:

    The frictional force can be calculated using the coefficient of friction, μ: Fextfriction=extμR=0.4imes25.4810.19extNF_{ ext{friction}} = ext{μ}R = 0.4 imes 25.48 ≈ 10.19 ext{ N}

    Thus, the frictional force acting on P as it moves up the plane is approximately 10 N.

Step 2

(b) the distance moved by P up the plane before P comes to instantaneous rest

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Answer

To find the distance moved by P up the plane, we can use the equations of motion.

  1. Acceleration Calculation:

    First, we need to find the net force acting on P while moving up the incline. The net force is given by: Fextnet=FextappliedFextfrictionWextparallelF_{ ext{net}} = F_{ ext{applied}} - F_{ ext{friction}} - W_{ ext{parallel}}

    The applied force here is the initial thrust exerted by P, which can be derived from its initial speed. Setting this in terms of acceleration, we find:

    The acceleration 'a' can be derived from: FFextfrictionWextparallel=maF - F_{ ext{friction}} - W_{ ext{parallel}} = ma Rearranging gives: a = rac{6 - 10.19 - 14.715}{3} ≈ - 8.3 ext{ m/s}^2

  2. Distance Calculation:

    We can use the equation of motion: v2=u2+2asv^2 = u^2 + 2as

    Given that the final speed v = 0 m/s, initial speed u = 6 m/s, we plug in the values:

    0=62+2imes(8.3)imess0 = 6^2 + 2 imes (-8.3) imes s 0=3616.6s0 = 36 - 16.6s Thus, solving for s, we have: s = rac{36}{16.6} ≈ 2.17 ext{ m}

    Therefore, the distance moved by P up the plane before it comes to instantaneous rest is approximately 2.17 m.

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