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].

Kerala Engineering Entrance 2009 Multiple Choice Questions on Work and Energy

In the KEAM (Engineering ) 2009 question paper three questions were included from the section ‘work, energy and power’. Here are those questions with solution:

(1) A particle is acted upon by a force F which varies with position x as shown in figure. If the particle at x = 0 has kinetic energy of 25 J, then the kinetic energy of the particle at x = 16 m is

(a) 45 J

(b) 30 J

(c) 70 J

(d) 135 J

(e) 20 J

The work done by a variable force acting along the direction of displacement can be found by drawing a force-displacement graph. The area under this graph gives the work done. Since the force has positive and negative values, the area also has positive (+50 units) and negative (– 30 units) values. The net area is +20 units and hence the gain in kinnetic energy during the movement from position x = 0 to the position x = 16 m is 20 J.

The particle has kinetic energy of 25 J at position x = 0. Therefore the kinetic energy at position x = 16 m is 25 + 20 = 45 J.

(2) Two springs P and Q of force constants k_P and k_Q [k_Q = k_P/2] are stretched by applying force of equal magnitude. If the energy stored in in Q is E, then the energy stored in P is

(a) E

(b) 2 E

(c) E/8

(d) E/4

(e) E/2

The potential energy (U) of a spring stretched (by a force F) through distance x is given by

U = ½ kx² where k is the spring constant (force constant) given by k = F/x

This can be rewritten as

U = ½ (F²/k) since x = F/k

The nergy stored in P and Q are respectively given by U_P = ½ (F²/k_P) and U_Q = ½ (F²/k_Q).

Therefore, U_P/U_Q = k_Q/k_P = ½ as given in the question.

This gives U_P = U_Q/2 = E/2.

(3) A rod of mass m and length l is made to stand at an angle of 60º with the vertical. Potential energy of this rod in this position is

(a) mgl

(b) mgl/2

(c) mgl/3

(d) mgl/4

(e) mgl/√2

When the rod is kept inclined at an angle of 60º with the vertical, its centre of gravity is raised (from the ground level) by a height h = (l/2)cos 60º = l/4.

The gravitational potential energy of the rod in this position is mgh = mgl/4.

You will find many useful multiple choice questions on work, energy and power at physicsplus and at AP Physics Resources