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Physics 101 Homework 8 Due February 7 1. A stone is dropped from a height 20 m above the ground. If the initial speed is zero, find speed of the stone just before it will reach the ground. (Neglect air resistance.) Solution: From the conservation of mechanical energy or from the work kinetic energy principle follows: mv 2 v 20m / s v 29.8m / s 2 20m 19.8m / s mgh v 2 gh 2 2. A stone is dropped from a height 20 m above the ground with initial velocity 10 m/s directed down. a) Find speed of the stone just before it will reach the ground. (Neglect air resistance.) b) What would be your answer if initial velocity is directed up? Solution: a) From the conservation of mechanical energy or from the work kinetic energy principle mv22 mv12 follows: mgh v 2 2 gh v12 2 2 v2 2 9.8m / s 2 20m 10m / s 22.2m / s v2 22m / s 2 b) The answer would be the same as above because kinetic energy depends on v 2 . Recall that that a stone sent up returns to the same point with the same speed (however, with opposite direction of velocity). 3. A box of mass 5.0 kg is accelerated from rest by a force across frictionless floor at a rate of 2.0 m s 2 for 7.0 s. Find the net work done on the box. Solution: Since the acceleration of the box is constant, we can find the distance moved: 2 x x x0 v0t 12 at 2 0 12 2.0 m s 2 7 s 49 m . Then the work done in moving the box is W F x cos 0o max 5 kg 2.0 m s 2 49 m 4.9 10 J 2 4. (a) If the kinetic energy of an arrow is doubled, by what factor has its speed increased? (b) If its speed is doubled, by what factor does its kinetic energy increase? Solution: (a) Since K 12 mv 2 , then v 2 K m and so v K . Thus if the kinetic energy is doubled, the speed will be multiplied by a factor of 2 . (b) Since K 12 mv 2 , then K v 2 . Thus if the speed is doubled, the kinetic energy will be multiplied by a factor of 4 . 1 Physics 101 Homework 8 Due February 7 5. How much work is required to stop an electron m 9.1110 31 kg which is moving with a speed of 1.90 106 m s ? Solution: The work done on the electron is equal to the change in its kinetic energy. W K 12 mv22 12 mv12 0 12 9.11 1031 kg 1.90 106 m s 2 1.64 1018 J 6. The roller-coaster car shown in Fig. is dragged up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4. Solution: Since there are no dissipative forces present, the mechanical energy of the roller coaster will be conserved. Subscript 1 represents the coaster at point 1, etc. The height of point 2 is the zero location for gravitational PE. We have v1 0 and y1 35 m . Point 2: 1 2 mv12 mgy1 12 mv22 mgy2 ; y2 0 mgy1 12 mv22 v2 2 gy1 2 9.80 m s 2 Point 3: 1 2 35 m 26 m s mv12 mgy1 12 mv32 mgy3 ; y3 28 m mgy1 12 mv32 mgy3 v3 2 g y1 y3 2 9.80 m s 2 Point 4: 1 2 7 m 12 m s mv12 mgy1 12 mv42 mgy4 ; y4 15 m mgy1 12 mv42 mgy1 v4 2 g y1 y4 2 9.80 m s 2 20 m 20 m s 2