EAMCET (Medical) 2009 Questions on Work and Energy

The following questions which appeared in the EAMCET (Medical) 2009 question paper are worth noting:

(1) A block of mass ‘m’ is connected to one end of a spring of spring constant ‘k’. The other end of the spring is fixed to a rigid support. If the mass is released slowly so that the total energy of the system is then constituted by only the potential energy, then ‘d’ is the maximum extension of the spring. Instead, if the mass is released suddenly from the same initial position, the maximum extension of the spring now is (g = acceleration due to gravity)

(1) mg/k

(2) 2d

(3) mg/3k

(4) 4d

The mass m is suspended by means of the spring. Since the spring is extended through a distance d, we have

mg = kd so that k = mg/d

When the mass is suddenly released, suppose the spring extends through an additional distance x. The total extension then is d+x.

The spring mass system momentarily comes to rest in the condition of maximum extension and then tries to return to the initial extension of d, executing simple harmonic oscillations. In the condition of maximum extension (equal to d+x) the gravitational potential energy mg(d+x) of the mass is converted into elastic potential energy of the spring so that we have

mg(d+x) = (½) k(d+x)²

Or, mg(d+x) = ½ (mg/d)(d+x)² since k = mg/d

This gives 2 = (d+x)/d from which x = d

The total extension d+x is therefore equal to 2d [Option (2)]

(2) A particle is projected up from a point at an angle θ, with the horizontal direction. At any time ‘t’, if ‘p’ is its linear momentum, ‘y’ is the vertical displacement and ‘x’ is the horizontal displacement, the graph among the following, which does not represent the variation of kinetic energy of the projectile is

(1) Graph (A)

(2) Graph (B)

(3) Graph (C)

(4) Graph (D)

The kinetic energy of a projectile has to decrease with the increase in its vertical displacement since its gravitational potential energy increases at the cost of its kinetic energy. Therefore graph (A) is incorrect.

[Graphs (B) and (C) are correct since the kinetic energy decreases with the increase in the horizontal displacement x, becomes a minimum at half the horizontal range (corresponding to the maximum height) and then increases. Graph (D) also is correct since the kinetic energy k is given by

k = p²/2m where p is the linear momentum and m is the mass of the particle.

Therefore, k is directly proportional to p², yielding a straight line graph].

Spring Constant and Potential Energy of Spring

Spring constant (force constant of a spring) is the force required for unit extension (or contraction) in a spring. Suppose a spring is cut into two pieces of equal length. Will the spring constant change? Don’t be doubtful. The spring constant will be doubled. If you cut a spring of spring constant ‘k’ into ‘n’ pieces of equal length, the spring constant of each piece will be nk.
Springs may appear in series and parallel combinations in certain questions, as for example, in problems involving the period of oscillation of a spring-mass system. If you have springs of constants k₁, k₂, k₃….etc. in series, the net spring constant ‘k’ of the combination is given by the reciprocal relation, 1/k = 1/k₁ + 1/k₂ + 1/k₃ + . . . . . . .etc.
If you have springs in parallel, the net spring constant ‘k’of the combination is given by k = k₁+ k₂+ k₃+ . . . . .etc.
Often you will encounter questions involving the potential energy of a spring. You should remember that the potential energy of a spring of constant ‘k’, stretched (or contracted) through a distance ‘x’ is (½) kx². Consider the following MCQ:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, its potential energy will be
(a) 3U (b) 6U (c) 9U (d) U (e) 36U
Since the potential energy is ½ kx², it follows that the P.E. is directly proportional to the square of the stretch (extension). The extension being 3 times, the P.E. must be 9 times. So, the correct option is (c).
Suppose the above question is asked in the following modified form:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, the increase in its potential energy will be
(a) 4U (b) 6U (c) 9U (d) 8U (e) 36U
Since the difference between the initial and final potential energies is required in this problem, the answer is 9U-U = 8U.
Now, consider the following question:
The tension in a spring of spring constant k is T. The potential energy of the spring is
(a) T²/k² (b) T²/k (c) 2T²/k (d) T²/2k (e) 2T²/k²
Potential energy of a spring, as you know, is (½)kx². But k=T/x from which x=T/k. On substituting this value of x in the expression for potential energy, we obtain option (d) as the answer.
Now, consider the following M.C.Q. which appeared in IIT screening 2002question paper:
An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially un-stretched. Then the maximum extension in the spring is
(a) 4Mg/k (b) 2Mg/k (c)Mg/k (d) Mg/2k
When the mass is released, its gravitational potential energy is decreased and the elastic potential energy in the spring is increased by an equal amount. If ‘x’ is the maximum extension produced (when the mass reaches the lowest position), we have Mgx = (½) kx², from which x = 2Mg/k. [Option (b)]