Work Energy Power Question

Document Title

1. An apartment block receives water from a nearby reservoir. A pump is necessary to lift the water into a storage tank at the top of the building. How much extra energy must the pump supply to each one kilogram of water, in order to lift it into the tank? (gravitational field strength is 10 N/kg)

              

(A) 100J    (B) 400J    (C) 500J    (D) 600J

There are two ways to analyze this question:
(1) By principle of conversation of energy, GPE at the surface of reservoir is same as GPE at 40m height of building. As water is flowing continuously, the water is able to flow up to 40m height of the building itself based on conservation of energy. Thus, the extra energy required is the GPE for 10m height.
(2) From the concept of manometer, the liquid level at both arms are the same when experiencing same pressure. So, the liquid at tank is able to reach 40m height of building itself. Thus, extra energy required is the GPE for 10m height.

Answer:
GPE = mgh = 1 * 10 * 10 = 100J

 Comment: a very common mistake is to simply use 50m as the height to calculate GPE.

2. The diagram below shows a boy of 65.0kg using a ‘chin-up’ bar to lift himself up.

                                    

Calculate
(a) the amount of work the boy has to do to complete one ‘chin-up’.
(b) the initial speed of the boy when lift himself up to complete one 'chin-up'.

Common mistakes:
(1) Forget to convert the unit from cm to m;
(2) Use wrong height to do the calculation.

(a) Analysis:
For a complete 'chin-up', we only need to lift our head above the bar. Thus, the correct height is 50cm, not the full height 150cm.
work done = GPE = mgh
                  = 65 * 10 * 0.5 = 325J
(b) From conservation of energy, KE = GPE = 325J
                325 = 0.5mv2
                325 = 0.5*65v2
                     v = 3.16m/s

3. A falling metal hammer is used to drive a hollow steel post into the ground as shown in the Figure below. The hammer is lifted by an electric motor and it then falls freely to hit the baseplate.

                  

(a) State the energy conversions from the time the hammer is released from its highest position till after impact.
(b) The metal hammer has a mass of 1500 kg and it is released from 3.2m above the baseplate. Calculate
      (i) the gravitational potential energy of the hammer
      (ii) the speed of the hammer as it hits the baseplate
(c) In raising the hammer, explain why the electric motor uses more energy than that stated in b(i).
(d) Assume the frictional force in the machinery is about 500 N, how much work is done to overcome this friction by lifting the hammer to a height of 3.20m?

(a) As the hammer falls, the gravitational potential energy is converted to kinetic energy.
Upon impact with the base plate, the kinetic energy is converted to sound energy, and heat and mechanical energy to push the base plate deeper into the ground.
(b)(i) GPE = mgh = 1500*10*3.2 = 48000J
      (ii) Assume no energy is lost to the surrounding,
                  KE = GPE = 48000J
                  48000 = 0.5mv2
                              v = 8.0m/s
(c) More energy is needed to overcome friction in the machinery that is used to lift the hammer.
(d) WD = F * d = 500 * 3.2 = 1600 J

4. The Figure below shows a roller coaster with a mass of 900 kg coasting on a track. When it reaches point A, it is moving at 5 m/s. Assume there is no friction and air resistance.

                  

(a) Calculate for the roller coaster,
      (i) its kinetic energy at point A.
      (ii) its velocity at point B.
      (iii) the minimum energy required for it to move from point A to point C.
(b) Explain if the roller coaster will reach point C if its velocity is 5 m/s at point A.
(c) Describe the energy changes which take place when the roller coaster is moving from point A to point C.
(d) Suggest an alternative design so that it will be more energy efficient.

(a)(i) KE = 0.5mv2
                = 0.5 * 900 * 52
                = 11250 J = 11300 J (3sf)
(ii) Comment: A common mistake is not to include KE at A into the calculation as many just treat GPE at A = KE at B and then find the velocity at B.
Analysis:
At A, there are both KE and GPE;
At B, there is only KE.
So, by conservation of energy,
KEA + GPEA = KEB
11250 + 900 * 10 * 5 = 0.5 * 900 * vB2
            vB = 11.2 m/s (3sf)
(iii) Analysis:
Since the question assumes there is no friction and air resistance, In order for coaster to move from A to C, it must possess enough energy to overcome the difference in GPE at A and C first. Thus, the minimum energy required is the difference in the GPE at A and C.
So, min. energy = change in GPE
                          = mg(hC – hA)
                          = 900 * 10 * (8 – 5) = 27000 J
(b) No as the KE at point A is less than the minimum energy required to move from A to C , therefore, it is unable to reach point C with a velocity of 5 m/s at point A.
(c) At point A, it has a GPE and KE.
From A to B, it is increasing in speed thus gain in KE; decrease in height thus lose in GPE.
From B to C, it is decreasing in speed thus lose in KE; increase in height thus gain in GPE.
(d) It is more feasible to design each successive hill lower so that the coaster will be able to move over each peak without supplying extra energy.


Excel@Physics