Work, Power & Energy Solved Problems

WORK, POWER & ENERGY Page # 8 SOLVED EXAMPLE Ex.1 Find the angle between force F ^ ^ ^ ^ ^ ^ unit and displacement d (3 i  4 j  5 k) = = (5 i  4 j  3 k) unit. Also find the projection of F on d. Ans. F.d = Fx dx Fy dy Fz dz = 3 (5) + 4 (4) + (– 3) (3) = 16 unit Hence F.d = F d cos= 16 unit Now F.F = F2  F2x  F2y  F2z = 9 + 16 + 25 = 50 unit and d.d = d2 = dx2  dy2  dz2 = 25 + 16 + 9 = 50 unit  cos  =  = cos –1 16 50 50  16  0.32 50 0.32 Ex.2 It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter isknown to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass1.00 g falling from a height 1.00 km. It hits the ground with a speed of 50.0 m s-1. (a) What is the work done by the gravitational force ? What is the work done by the unknown resistive force? Ans. (a) The change in kinetic energy of the drop is K = 1 m2 – 0 2 1 × 10–3 × 50 × 50 = 1.25 J 2 where we have assumed that the drop is initially at rest. Assuming that g is a constant with a value 10 m/s2, the work done by the gravitational force is, Wg = mgh = 10-3 × 10 × 103 = 10.0 J (b) From the work-energy theorem K = Wg + Wr where Wr is the work done by the resistive force on the raindrop. Thus Wr = K – Wg = 1.25 –10 = – 8.75 J is negative. done by the stopping (frictional) force on the cycle due to the road. (a) The stopping force and the displacement make an angle of 180º (rad) with each other. Thus, work done by the road, Wr = Fd cos = 200 10 cos = – 2000 J It is this negative work that brings the cycle to a halt in accordance with WE theorem. (b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude is 200 N. However, the road undergoes no displacement. Thus, work done by cycle on the road is zero. .. Ex.4 In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 m s-1 (see Table 6.2) on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet ? Ans. The initial kinetic energy of the bullet is mv /2 = 1000 J. It has a final kinetic energy of 0.11000 = 100 J. If v is the emergent speed of the bullet, 2 f 1 m2f  100 J f  2 2  100 J =63.2 ms–1 0.05kg Ex.5 A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The total distance through which the trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N. Calculate the work done by the two forces over 20 m. Ans. = Ex.3 A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle ? (b) How much work does the cycle do on the road ? Ans. Work done on the cycle by the road is the work \ The plot of the applied force is shown in Fig. At x = 20 m, F = 50 N ( 0). We are given that the frictional force f is |f|= 50 N. It opposes motion and acts in a direction opposite to F. It is therefore, shown on the negative side of the force axis. The work done by the woman is WF area of the rectangle ABCD + area of the trapezium CEID 1 WF 100 ×10 + (100 + 50) × 10 2 = 1000 + 750 =1750 J The work done by the frictional force is WF area of the rectangle AGHI WORK, POWER & ENERGY Page # 9 Wf (–50) × 20 = – 1000 J The area on the negative side of the force axis has a negative sign. Ex.6 A block of mass m = 1 kg, moving on a horizontal surface with speed vi = 2 ms–1 enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force Fr on the block in this range is inversely proportional to x over this k range, qF1 = for 0.1 < x < 2.01 m = 0 for x < x 0.1m and x > 2.01 m where k = 0.5 J. What is the final kinetic energy and speed vf of the block as it crosses this patch ? xf Ans. From Eq. kf – ki = =  xi 2.01 F dx K = K + f i  0 .1 (k ) dx x 1 1 m2t  k ln(x) |20..01 m2t  k ln(2.01 / 0.1) 1 = 2 2 = 2 – 0.5 ln (20.1) = 2 – 1.5 = 0.5 J f = 2K f / m = 1 ms–1 Ex.7 A bob of mass m is suspended by a light string of length L . It is imparted a horizontal velocity vo at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for (i) vo; (ii) the speeds at points B and C; (iii) the ratio of the kinetic energies (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C. E = 1 2 mv0 2 m20 [Newton’s Second Law] L where TA is the tension in the string at A. At the TA  mg  highest point C, the string slackens, as the tension in the string (TC) becomes zero. Thus, at C 1 m2c  2mgL 2 where vC is the speed at C. E E .....(1) 5 mgL 2 ....(2) Equating this to the energy at A or (ii) 0 = 5 m 2 mgL  0 2 2 5gL It is clear from C = gL At B, the energy is 1 mB2  mgL 2 Equating this to the energy at A and employing the E result from (i), namely v02 = 5gL  1 1 mB2  mgL  m20 2 2 B = 3gL (iii) The ratio of the kinetic energies at B and C is : = 5 mgL 2 1 mB2 KB 3 2   1 KC 1 2 mC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. Ans. (i) There are two external forces on the bob : gravity and the tension (T ) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the system is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A :