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A box of mass 2 kg is pulled up a rough plane face by means of a light rope - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

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A box of mass 2 kg is pulled up a rough plane face by means of a light rope. The plane is inclined at an angle of 20° to the horizontal, as shown in Figure 2. The ro... show full transcript

Worked Solution & Example Answer:A box of mass 2 kg is pulled up a rough plane face by means of a light rope - Edexcel - A-Level Maths Mechanics - Question 4 - 2005 - Paper 1

Step 1

a) the normal reaction of the plane on the box

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Answer

To find the normal reaction (R) of the plane on the box, we can use the following formula based on the forces acting upon the box along the direction perpendicular to the plane:

  1. The weight of the box can be resolved into two components: parallel (2gimesextsin(20exto)2g imes ext{sin}(20^ ext{o})) and perpendicular (2gimesextcos(20exto)2g imes ext{cos}(20^ ext{o})) to the plane.

  2. The equation for the normal reaction is given by:

    R=2gimesextcos(20exto)R = 2g imes ext{cos}(20^ ext{o})

    Substituting the known values (where gext(accelerationduetogravity)extisapproximately9.8extm/s2g ext{ (acceleration due to gravity)} ext{ is approximately } 9.8 ext{ m/s}^2):

    R=2imes9.8imesextcos(20exto)R = 2 imes 9.8 imes ext{cos}(20^ ext{o})

    Calculating this gives: Rextapproximately18.4extNR ext{ approximately } 18.4 ext{ N}

    Therefore, the normal reaction of the plane on the box is approximately 18 N.

Step 2

b) the acceleration of the box

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Answer

To determine the acceleration of the box, we analyze the forces acting parallel to the plane:

  1. The force parallel to the plane is given by:

    Rext(//toplane)=182gimesextsin(20exto)F=2aR_{ ext{(// to plane)}} = 18 - 2g imes ext{sin}(20^ ext{o}) - F = 2a

    where F is the frictional force, which can be calculated as:

    F=0.6RF = 0.6R

  2. Substituting the value of R obtained from part (a):

    F=0.6imes18extN=10.8extNF = 0.6 imes 18 ext{ N} = 10.8 ext{ N}

  3. Now we can substitute F back into the force equation:

    182imes9.8imesextsin(20exto)10.8=2a18 - 2 imes 9.8 imes ext{sin}(20^ ext{o}) - 10.8 = 2a

  4. Solving this equation will yield:

    a = rac{(18 - 10.8 - 2g imes ext{sin}(20^ ext{o}}{2}

    Finally, calculating will lead to: aextapproximately0.12extm/s2a ext{ approximately } 0.12 ext{ m/s}^2.

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