View Full Version : help with a physics problem
dell98
2004 April 15th, 11:55
I keep coming up with different answers for this problem.
A crate with a mass of 150 kg is placed on soft ground that sinks under the weight of the box at 0.5 m/s2. How much force is the box exerting on the ground? (g = 10 m/s2).
I get 750N or 1425N depending on how I look at the question.
Help?
Cin.
[Edited on 4-15-2004 by dell98]
Ryth
2004 April 21st, 14:32
Regarding your problem- is there an extra weight acting down on the crate, because normally the normal force acting on an object is 9.8 N or 9.8 ms2. The equation should be F=mg, but you have an accelleration of .5 ms2. I do not understand where the 10 ms2 came from, but F=mga=750 N is wrong as far as I can tell. I am assuming that the acceleration does not matter and that the F=(150kg)(9.8 ms2) = 1470 N is the correct answer. If, on the otherhand, the force of g is in fact 10 ms2 then the f= 1500 N. Where did you get the figure 1425 N?
I have a problem of my own as well. Maybe you can help?
I am trying to determine the electrical energy used by a kettle. I believe the equation used is E(h)=E(l)+w. The only problem is that I do not no what to substitute for E(l)(Quantity of thermal energy flowing out at the lower temperature). Now, assuming that E(l) can be calculated by E(l)=ml(l being the latency heat) then I do not have a problem, but this does not sit well with me. The reason being that there is no change of state in the water(it is not evaporated into steam and it is not melted from ice). The only other formula I can think of using is mc-deltaT(c=heat capacity), but I am not sure.
The figures known are mass of water= 1kg, initial temperature= 80 degrees c, final temperature=59 degrees c, temperature change(hot water)=21 degrees c, temperature change(melted ice water)=59 degrees c(original temperature= 0 degrees c), time for heating=180s, and power of kettle=500w.
needless to say any advice would be of great help.
:puzzled::wall:
[Edited on 4-21-2004 by Ryth]
[Edited on 4-22-2004 by Ryth]
[Edited on 4-22-2004 by Ryth]
PritamDamania
2004 June 10th, 03:21
The problem is a simmple newtons law problem. Let the force exerted by the ground be F. Then
150*10 - F = 150*0.5
Thus F=1425
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