here's a set of 5 questions: Work and Energy 1. The problem statement, all variables and given/known data 1. While traveling to school at 27 m/s your car runs out of gas 15 km from the nearest gas station. If the station is 16 m above your current elevation, how fast will the car be going when it reaches the gas station? Ignore friction. 2. A lead ball is traveling at 275 m/s when it strikes a steel plate and comes to a stop. If all its kinetic energy is converted into heat and none of the energy leaves the bullet, what is the bullet’s change in temperature? 3. A 637 gram sample of water at 92 C is mixed with 843 grams of water at 27 C. Assume no heat loss to the surroundings. What is the final temperature of the mixture? 4. Tommy throws his car keys straight up in to the air from height of 1.5 m above the ground. His keys strike the ground at a speed of 7.7 m/s. Ignoring friction, how fast did Tommy throw his keys? 5. A 1.5 kg bass is hooked by a fisherman. The fisherman plays the bass by allowing the fish to swim off at 2.1 m/s before braking his reel and stopping the bass in 37 cm. How much tension is exerted on the line? Assume the fish is neutrally buoyant. 2. Relevant equations i know that: (Clead = 125 J/kg•K) (Cwater = 4180 J/kgK) 3. The attempt at a solution I did all of these in my head/on a calculator, but i'm not sure how any of the equations work or what they are. I've always been able to get the answer without any knowledge of how i did it. Help?? 20.8 m/s 302.5 C 54.98 C 5.47 m/s 8.94 N
Please read this https://www.physicsforums.com/showthread.php?t=94379 If you can't be troubled to write out your solution, we are not going to do the problems just to verify your answers. Why do you think you need help? I'll pick one. How did you do #2?
well, i've already confirmed the answers with a key the teacher has posted online. but as to getting to the answer, i just performed "random" functions in the calculator until i got to the answers that seemed correct. i have a basic idea... but, since you ask: #2 equations: Q=mcT U=Q-W U=mgh 1/2 mv^2=W W + U = mcT 1/2 mv^2 + mgh = mcT (v^2 + h) / 2c =T assuming the displacement (h) is zero, that works?
I think you should spend, oh, a few months reading your textbook. You're never going to get anywhere with this "technique." - Warren
Q=mcT <== Is any heat added to the lead in this problem? What is T in this equation? What is c? What value did you use in your computation? U=Q-W <== What is U and how is it related to T? Is any work done on by the lead (assume the plate does not move). Is any work done on the lead? What is the significance of this for the sign of W in this equation? How do your answers connect work done to changing the temperature? U=mgh <== Is this U the same as the previous U? 1/2 mv^2=W <== Is this the same W as in the U = Q-W equation? W + U = mcT <== What if Q = 0? What does this mean? (v^2 + h) / 2c =T <== If h were not zero this would not have worked. You dropped the g, and lost a factor of 2 Try to answer the questions. They are important to your understanding of how the correct final equation (v^2)/2c = ΔT gives you the answer
thanks for the help - one last question. after looking over all the other questions (and reading the book), i figured them out. but i am completely lost on #3. like, i have NO idea what to do. could someone help me reach a working equation?
The correct version of Q=mcT and the fact that heat lost by the water that cools = heat gained by the water that warms up.