Setting an example is not the main means of influencing others; it is the only means.
– Albert Einstein
Let us discuss a few multiple choice questions on work and energy.
(1) A small sphere of mass 20 g is projected vertically up with a velocity of 10 ms–1. If air resistance is negligible, what is the total work done by gravity during the upward trip of the sphere?
(a) 10 J
(b) –10 J
(c) – 1 J
(d) 1 J
(e) 9.8 J
When the sphere rises up its kinetic energy goes on decreasing because of the work done by the gravitational force against the motion of the sphere. (The gravitational potential energy of the sphere goes on increasing by an equal amount). The work done by gravity is evidently negative. When the sphere reaches the maximum height the entire kinetic energy gets converted into gravitational potential energy. The total work done by gravity during the the upward trip of the sphere is numerically equal to the initial kinetic energy (½ mv2) of the sphere but its sign is negative.
Therefore, the answer is – ½ mv2 = – ½ ×0.020×102 = –1 J.
(2) A world class athlete covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range
(a) 2 × 105 J to 5 × 105 J
(b) 2 × 104 J to 5 × 104 J
(c) 2 × 103 J to 5 × 103 J
(d) 200 J to 500 J
(e) 20 J to 50 J
This question has appeared in various entrance tests with slight differences in the wording and in the options. Last year it appeared in the All India Engineering Entrance Examination.
In the 100 m dash the velocity v is almost constant so that we have
v = 100 m/10 s = 10 ms–1
We may take the mass of the athlete to be 70 kg to 80 kg. Let us use 80 kg.
His kinetic energy will be ½ mv2 = ½ ×80×102 = 4000 J so that the correct option is (c).
Questions (3) and (4) appeared in Kerala Engineering Entrance Examination 2008 question paper.
(3) Two bodies A and B have masses 20 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times tA and tB, then the ratio tA/ tB is
(a) ½
(b) 2
(c) 2/5
(d) 5/6
(e) 1/5
The concept of impulse will be very useful here. The impulse received by A and B in the times tA and tB are respectively F tA and F tB where F is the force acting on them (4 kg wt. here). But impulse is the change in momentum so that the momenta acquired by A and B are F tA and F tB.
Since the kinetic energy is p2/2m where p is the momentum, we have
(F tA)2/2mA = (F tB)2/2mB
Therefore, tA/ tB = √( mA/ mB) = √(20/5) = 2.
(4) A particle acted upon by constant forces 4i + j – 3k and 3i + j – k is displaced from the point i + 2j + 3k to the point 5i + 4j + k. The total work done by the forces in SI units is
(a) 20
(b) 40
(c) 50
(d) 30
(e) 35
The resutant force (F) on the particle is the sum of the forces given by
F = (4i + j – 3k) + (3i + j – k) = 7i + 2j – 4k.
The displacement (s) of the particle is given by
s = (5i + 4j + k) – (i + 2j + 3k) = (4i + 2j – 2k)
The work done (W) is given by
W = F.s = (7i + 2j – 4k) . (4i + 2j – 2k) = 28 + 4 + 8 = 40.
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