Exam-style Questions: Forces

 

  1. This question is about driving poles into the ground.

    Fig. 8.1 shows a 220 kg mass held in osition 5.0 m above the top of a rigid, cylindrical pole. The lower end of the vertical pole is resting on the ground.

    Copyright S-cool

    When released, the mass drops freely from the rest under gravity and strikes the top of the pole.

    a) Describe the energy taking place from the moment the mass falls until it strikes the top of the pole.

    (2 Marks)

    b) Show that the speed of the mass is about 10 m s-1 when it strikes the top of the pole.

    g = 9.8 N kg-1

    (2 Marks)

    c) In bringing the moving mass to reat on top of the vertical pole, the pole is pushed down into the ground. The depth of penetration of the pole into the ground is 0.4m.

    Show that the average force exerted by the pole on the mass is about 27 kN.

    (2 Marks)

    (Marks available: 6)

  2. A boat of mass m = 2000 kg is moving with velocity u = 6.0 m s-1 towards a dock. The boat is stopped by a constant braking force F = 300 N.

    Complete the following table in order to calculate the time, t, and the distance, d, the boat takes to stop.

    Starting from force = mass x acceleration and the appropriate equation of motion show that Ft = mu. Starting from force = mass x acceleration and the appropriate equation of motion show that Fd = ½mu2

    Calculate the boat's initial momentum.

    momentum = .................. N s

    Calculate the boat's initial kinetic energy

    kinetic energy = ................ N s

    Calculate the time the boat takes to stop

    time = .................. s

    Calculate the distance the boat takes to stop.

    distance = .................. m

     

    (8 Marks)

Answer

 

Answer outline and marking scheme for question:

 

  1. a) gravitational potential energy to kinetic energy (and internal energy in air)

    (2 Marks)

    b) v = √(2 x 9.8 x 5) (1 Mark) = 9.9 m s-1 (1 Mark) (explicit)

    (2 Marks)

    c) ½ mv2 ≈ F x 0.4 (1 Mark) F ≈ x 27 500 N (1 Mark)

    or

    a ≈ (10)2/(2 x 0.4) = 125 m s-2 F ≈220a = 27 500 N

    or

    mgh = F x 0.4 F = 26 950 N

    (2 Marks)

    (Marks available: 6)

  2. F = ma

    v = 0 (1 Mark)

    F = (-) m u / t (1 Mark)

    Hence Ft = mv (0 Marks)

    v2 = u2 + 2ad

    a = (-) u2/2d (1 Mark)

    F = m u2/2d (1 Mark)

    Hence Fd = ½ mv2

    mv = 2000 kg x 6 m s-1

    = 12 000 N s (1 Mark)

    ½ mv2 = ½ x 2000 x 62

    = 36 000 N m OR J (1 Mark)

    12 000 Ns / 300 N

    = 40 s (1 Mark)

    36 000 N m / 300 N

    =120 m (1 Mark)

     

    (8 Marks)