View Full Version : Work and Energy Related Problem
Alethia
Nov23-03, 07:04 PM
Okay, I have this problem that I've been having trouble with. It seems like a simple problem, but I keep getting the wrong answer and I don't know where I'm making the error. Maybe one of you can help me target the problem. =) The problem is as follows:
If an automobile engine delivers 50.0 hp of power, how much time will it take for the engine to do 6.40 x 105 J of work? Well, firstly I converted the 50.0 hp into watts by multiplying it by 746 (because one hp equals 746 watts). I got 37300 watts. Then using the formula P=\frac{W}{\Delta t}}, I plug in 6.40 x 105 J as P(ower) and 6.40 x 105 as W(ork). When I calculated it, I get .058 seconds for t. However, the correct answer is 17.2 seconds. Can anybody tell me where I'm going wrong and waht I need to do?
You must be plugging things in wrong. Try again (you have everything right):
ΔT = W/P
Alethia
Nov23-03, 07:36 PM
Ohhh... haha. Whoops, I was just making a calculation error. Thanks! Okay I get it. I have another question though. In the following problem: A 1.0 x 103 kg elevator carries a maximum load of 800.0 kg. A constant frictional force of 4.0 x 103 N retards the elevator's motion upward. What minimum power, in kilowatts, must the motor deliver to lift the fully loaded elevator at a constant speed of m/s? Okay, for this problem, I'm not quite sure how to approach it. Initally, I thought I had to simply use the formula P=\frac{Fd}{\Delta t}} to solve it by pluggin in 4.0 x 103 N as F and 3.00 m as d, and jsut put in one second for t. However, when I calculated it, it didn't come out to the correct solution so I'm assuming that this approach was not right. How then, would I solve this problem?
You know that P = F(Δd/Δt)
Δd/Δt is equal to velocity, so P can also be expressed
P = Fv
The force the lift motor has to overcome when the lift is fully loaded is
F = (1000 + 800)g + 4000
So
P = (1800g + 4000)v
You didn't put the value of v in your question, but you mentioned that the lift moves 3m in one second, so
P = 3(1800g + 4000)
Originally posted by Alethia
However, when I calculated it, it didn't come out to the correct solution so I'm assuming that this approach was not right. How then, would I solve this problem?
You forgot that the motor has to work against gravity, not just the friction.
Alethia
Nov23-03, 08:26 PM
What do you mean when you refer to 'g'?
The constant g is the acceleration due to gravity.
Alethia
Nov23-03, 08:31 PM
Originally posted by AD
The constant g is the acceleration due to gravity. So then would I have to multiply 1800 by 9.81 m/s2?
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