Y12 Power Answers

1. Find the work done to lift a mass of 10 kg through a height of 5 m. If this takes 15s what is the power?

W= Fx = (10kg)(9.8Nkg^-1) (5m) = 490J

 

P = W/t = 490 J / 15s = 32.67 W

 

 

2         A mass of 5 kg is pulled along a surface at a constant speed. Frictional force between the mass and the surface is 4N what work is done in moving the mass 2 m along the surface? Find the power if this takes 2 s.

 

W = Fx = (4N) (2m) = 8J

 

P = W/t =  8J/ 2s = 4 W

 

3        A girl of mass 50 kg runs up a flight of steps 4.5 m high in 5 s. What power does she develop?

W = Fx             F = 50kg x 9.8 = 490.5 N                     x = 4.5m

W = 490.5 x 4.5 = 2207.25 J

P = W/t = 2207.25 J / 5s = 442w

 

4        2 x 10⁶ kg of water per second flow over a large waterfall of height 50 m.

W = Fx                   F = 2 x 10⁶ kg x 9.8 = 1.96 x10⁺⁷N      x=50m

W= (1.96 x10⁺⁷N) (50m) = 9.8 x10⁺⁸ J

What power is available?

As 1W = I J per second the water falling over the fall transfers 9.8 x10⁺⁸ J of GPE per second the available power is 9.8 x10⁺⁸ W (ignoring losses due to friction)

 

5        The first-stage rocket motors of Saturn 5 burnt fuel at the rate of 13600 kg per second. The work done by the pumps to drive the fuel into the combustion chambers is the same as that necessary to lift the fuel through 1680 m.

W= Fx                    F = 13600kg x 9.8= 1.3328 x 10⁺⁵ N    x = 1680m

W = (1.3328x 10⁺⁵ ) (1680) = 2.24 x 10⁺⁸ J

What power is needed to do this? (Your answer is about twice the engine power of the largest ocean liner.)

As this mass of fuel is burnt in 1 second then 2.24 x 10⁺⁸ J of work is done in 1 second so the power is 2.24 x 10⁺⁸ W

 

6        A gas turbine locomotive developing 6330 kW pulls a train at a steady velocity of 108 kmh^{- 1}. What is the total force in Newtons resisting the motion?

At a steady velocity all work done by the engine is against friction or gravity.

The engine does 6.33 x10⁶ J of work every second

As W = Fx the force exerted by the engine is equal Work done per second divided by the distance travelled per second

F = W/x            108 kmh^-1 = (108000m)/(60 min x 60 sec) = 30 ms^-1

F = 6.33 x10⁶ J/ 30m = 211 000N = 2.11 x 10⁵ N

Nov Assesment Mark Scheme

SHM test

1.       The acceleration of the oscillator is directly proportional to the                              B1
displacement (the term displacement to be included and spelled correctly
to gain the mark).
with the acceleration always directed to a fixed / equilibrium point                        B1

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  2.     (a)     arrow towards centre of planet                                                                         1

(b)     (i)      g = GM/R2                                                                                               1

(ii)     gS/gO = R2/25R2; gS = 40/25 (= 1.6 N kg–1)                                             2

(iii)    gC/gO = R2/16R2 giving gS = 40/16 (= 2.5 N kg–1)                                    1

(iv)    average g = (2.5 + 1.6)/2 = 2.(05) (1)
Δp.e. (= mgavR) = 3.0 × 103 × 2.05 × 2.0 × 107; = 1.2 × 1011 (J) (2)       3

  (c)   g = v2/r; = 4π2(5R)/T2 (2)
1.6 = 4 × 9.87 × 1.0 × 108/T2 giving T2 = 24.7 × 108 and T = 5.0 × 104 (s) (2) 4

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  3.     (a)     (i)      Fig. 1 : x and a in opposite directions/acceleration towards
equilibrium point/AW; (1)
Fig. 2 : proportional graph between x and a/AW (1)                                2
Figures not identified max. of 1 mark

  (ii)   a = 4π2f2x; 50 = 4π2f2.50 × 10–3; giving f2 = 25 and f = 5.0 Hz              3

  (iii)  cosine wave with initial amplitude 25 mm; decreasing amplitude; (2)
correct period of 0.2 s (for minimum of 2.5 periods); (1)                       3

  (b)   (i)      the acceleration towards A/centripetal acceleration or force;
is constant                                                                                               2

  (ii)   a = v2/r; so 50 = v2/10; v2 = 500 giving v = 22.4 m s–1                            3

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  4.     (a)     (i)      v = 2πrf = 2π × 0.015 × 50; = 4.7 (m s–1)                                                2

  (ii)   a = v2/r = 4.72/0.015; = 1.5 × 103 (m s–2) ecf(a)(i)                                   2

  (iii)  the belt tension is insufficient to provide the centripetal force; (1)
so the belt does not ‘grip’ the pulley/does not hold the belt against
the pulley/there is insufficient friction to pull/push/move the belt. (1)    2
alternative argument the belt does not ‘grip’ the pulley/there is
insufficient friction to pull/push/move the belt; because of its
inertia/insufficient to provide force for acceleration of (belt)-drum

  (b)   resonance occurs; when the natural frequency of vibration of the (1)
panel = rotational frequency of the motor (1)                                                   2

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shm assessed hw markscheme

1.       (a)     The resultant force is zero (WTTE)

For the first mark allow
-      sum of forces is zero,
-      upward force = downward force,
-      forces cancel each other
BUT do not allow forces are balanced

B1

          Forces are weight and force from the spring (allow tension)

Allow force of gravity for weight

B1

  (b)   (i)      acceleration is (directly) proportional to displacement

M1

          and is directed in the opposite direction to the displacement. (WTTE)

allow a = –(2πf)2 x, provided a and x are identified and –ve sign must be explained.
Do not allow “acceleration is prop to negative displacement for second mark.
Allow always towards the equilibrium position

A1

  (ii)   x = acos2πft  2πf = 7.85 (expressed in any form)

M1

          f = (7.85/2π) = 1.25 (1.249Hz)

Do not allow use of the fig to show T = 0.8s and hence
f = 1.25 Hz. This scores 0.

A1

  (iii)  correct substn in Vmax = (2πf)A  Vmax = 2π × 1.25 × 0.012

Many will forget to change 12 mm into 0.012m and have
v = 94 m s–1 this scores 1 mark.

C1

          Vmax = 0.094 m s–1

A1

  (c)   roughly sinusoidal graph of correct period ie 0.8s

B1

          90° out of phase with displacement graph (i.e. starts at origin
with -ve initial gradient)

B1

          maximum velocity correctly shown as 0.094 {allow ecf from (iii)}

B1

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  2.     (i)      1.      Measure the time t for N oscillations.                                                   M1
frequency f = N/t                                                                                  A1

2.       Measure the amplitude A of the oscillations using the ruler.                 M1
maximum speed is calculated using: vmax = (2πf)A                               A1

  (ii)   The maximum speed is doubled                                                                    B1
because the frequency is the same and vmax =  2 pi fA

(iii)    F = (–) kx and F = ma
Therefore ma = (–) kx                                                                                   M2
ω2 = k / m
T = 2Pi root k over m                                                                                              M1

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  3.     (a)     (i)      A motion in which the acceleration/force is proportional to the
displacement; (1)
directed towards the centre of oscillation/equilibrium position/AW
or a α -x or a = –ω2x or a = –4π2f2x; symbols must be identified (1)     2

  (ii)   T = 0.25 s or f = 1/T; f = 4 (Hz) (2)                                                         2

(iii)    a = –4π2f2A; = 4 × 9.87 × 16 × 0.005; = 3.2 (m s–2) ecf a(ii) (3)             3

  (b)   (i)      Resonance occurs at /close to the natural frequency of an oscillating (1)
object/system; caused by driving force (at this frequency); when (1)
maximum energy transfer between driver and driven/maximum
amplitude achieved (1)                                                                            3

3 marking points in any sensible order

  (ii)   1   reduced amplitude; as resonance frequency lower
     or resonance will occur at lower frequency; as greater
     inertia/reduced natural frequency/AW in terms of amplitude change (2)

          2   reduced amplitude; as resonance frequency higher
     or resonance will occur at a higher frequency; as larger restoring
     force/increased natural frequency/AW in terms of amplitude change (2)           4

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  4.     (a)     (i)      acceleration ∞ displacement; indication of restoring force by negative
sign/acc. in opp. direction to displacement/acc. towards origin/AW        2

  (ii)   linear graph through origin; negative gradient                                          2

  (b)   (i)      0.05 (m)                                                                                                  1

  (ii)   4π2f2 = a/A; = 12.5/0.05 = 250 so f = 2.5(1) Hz; T = 1/f (= 0.4 s)           3

  (c)    (i)      cosine wave; correct period of 0.4 s; correct amplitude of 0.05 m          3

  (ii)   0; 0.1/0.3/0.5/0.7/0.9 (s)                                                                          2

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