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 kP and kQ [kQ = kP/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 UP = ½ (F2/kP) and UQ = ½ (F2/kQ).
Therefore, UP/UQ = kQ/kP = ½ as given in the question.
This gives UP = UQ/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.
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