In the movie Eraser, the bad guys had a gun that would shoot a 5 g aluminum slug

Question

In the movie Eraser, the bad guys had a gun that would shoot a 5 g aluminum slug at one quarter the speed of light. Let's assume that the slug is acted on by a constant force over the length of the 1 m long barrel. Further assume that the person shooting the gun is capable of keeping the gun stationary as the slug is being fired. a) What work is done by the gun on the slug? b) To put this number into perspective, calculate how high you would have to lift a 1500kg car in a constant 1 g gravitational field in order to equal the work calculated in part a).

Explanation / Answer

FOLLOW THIS here are two methods of solving this problem. The first method involves using the equation W = F*d*cos(Theta) where F=20.8 N, d=0.636 m, and Theta=0 degrees. (The angle theta represents the angle betwee the force and the displacement vector; since the force is applied parallel to the incline, the angle is zero.) Substituting and solving yields W = F*d*cos(Theta) = (20.8 N)*(0.636 m)*cos(0) = 13.2 J. The second method is to recognize that the work done in pulling the cart along the incline at constant speed changes the potential energy of the cart. The work done equals the potential energy change. Thus, W=Delta PE = m*g*(delta h) = (3.00 kg)*(9.8 m/s/s)*(0.45 m) = 13.2 J Useful Web Links Definition and Mathematics of Work | Potential Energy [ #26 | #27 | #28 | #29 | #30 | #31 | #32 | #33 | #34 | #35 | #36 ] 28. Eddy, whose mass is 65.0-kg, climbs up the 1.60-meter high stairs in 1.20 s. Approximate Eddy's power rating. PSYW Answer: P = 849 Watts Eddy's power is found by dividing the work which he does by the time in which he does it. The work done in elevating his 65.0-kg mass up the stairs is determined using the equation W = F*d*cos(Theta) where F = m*g = 637 N (the weight of the 65.0 kg object), d =1.60 m and Theta = 0 degrees (the angle between the upward force and the upward displacement). Solving for W yields 1019.2 Joules. Now divide the work by the time to determine the power: P = W/t = (1019.2 J)/(1.20 s) = 849 Watts Useful Web Links Definition and Mathematics of Work | Power [ #26 | #27 | #28 | #29 | #30 | #31 | #32 | #33 | #34 | #35 | #36 ] 29. A 51.7-kg hiker ascends a 43.2-meter high hill at a constant speed of 1.20 m/s. If it takes 384 s to climb the hill, then determine ... . PSYW kinetic energy change of the hiker. the potential energy change of the hiker. the work done upon the hiker. the power delivered by the hiker. Answers: Delta KE = 0 J Delta PE = +21900 J W = +21900 J P = 57.0 Watts a. The speed of the hiker is constant so there is no change in kinetic energy - 0 J. b. The potential energy change can be found by subtracting the initial PE (0 J) from the final PE (m*g*hf). The final potential energy is 21888 J [from (51.7 kg)*(9.8 m/s/s)*(43.2 m)] and the initial potential energy is 0 J. So Delta PE = +21900 J (rounded from 21888 J). c. The work done upon the hiker can be found using the work-energy theorem. The equation reduces to Wnc = PEf (PEi = 0 J since the hiker starts on the ground; and KEi = KEf since the speed is constant; these two terms can be dropped from the equation since they are equal). The final potential energy is 21888 J [from (51.7 kg)*(9.8 m/s/s)*(43.2 m)]. So W = +21900 J (rounded from 21888 J). d. The power of the hiker can be found by dividing the work by the time. P=W/t=(21888 J)/(384 s) = 57.0 Watts Useful Web Links Potential Energy | Kinetic Energy | Definition and Mathematics of Work | Power [ #26 | #27 | #28 | #29 | #30 | #31 | #32 | #33 | #34 | #35 | #36 ] 30. An 878-kg car skids to a stop across a horizontal surface over a distance of 45.2 m. The average force acting upon the car is 7160 N. Determine ... . PSYW the work done upon the car. the initial kinetic energy of the car. the acceleration of the car. the initial velocity of the car. Answers: W = -324000 J KEi = +324000 J a = -8.16 m/s/s vi = 27.2 m/s a. The work done upon the car can be found using the equation W = F*d*cos(Theta) where F=7160 N, d=45.2 m, and Theta=180 degrees (the force is in the opposite direction as the displacement). Substituting and solving yields -323632 J (rounded to -324000 J). b. The initial kinetic energy can be found using the work-energy theorem. The equation reduces to KEi + Wnc = 0 (PEi and PEf = 0 J since the car is on the ground; and KEf = 0 J since the car is finally stopped). Rearrange the equation and it takes the form KEi = -Wnc . So KEi = +324000 J (rounded from +323632 J). c. The acceleration of the car can be found using Newton's second law of motion: Fnet = m*a The friction force is the net force (since the up and down forces balance) and the mass is 878 kg. Substituting and solving yields a = -8.16 m/s/s. d. The initial velocity of the car can be found using the KE equation: KE = 0.5*m*v2 where m=878 kg and KEi=323632 J. Substituting and solving for velocity (v) yields v = 27.2 m/s. (A kinematic equation could be also used to find the initial velocity.)

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