Calculating Work and Average Force in a Meteor Impact on Sand

AI Thread Summary
The discussion revolves around calculating the work done by sand to stop a meteor, its average force, and the thermal energy produced during the impact. Participants agree that the work done should equal the change in kinetic energy, calculated using the meteor's mass and velocity. However, they encounter discrepancies in their calculations, particularly with the final velocity and the resulting force. Despite using formulas for kinetic energy and force, the answers do not match the expected results from an external website, leading to confusion about the accuracy of their computations. The thread emphasizes the importance of correctly applying physics principles to solve the problem.
jbot2222
Messages
14
Reaction score
0
A meteor has a speed of 79.8 m/s when 787 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.19 m.
~How much work does the sand do to stop the meteor (mass = 556 kg)?
~What is the average force exerted by the sand on the meteor?
~How much thermal energy is produced?

A meteor falls to Earth and hits sand. The work done by the sand to stop it should just equal the change in kinetic energy. I found the final velocity before it hit the sand, and from that I found the decceleration of the meteor in the sand. That acceleration times the mass and multiplyed by the distance it took to stop should give the total work done by the sand as long as energy is conserved. But it doenst work!

As for the average force, that should be the mass of the meteor*the acceleration. that doest work either
 
Physics news on Phys.org
Hmm, how are you sure that these are wrong?
OK, KE = 1/2mv^2 and v at that low point = (79.8^2 + 19.6(787000))^0.5 = (15431568.04)^0.5 = 3928.30ms^-1
So KE = 4289968367.42J = W, is that what you got?

F = ma
0^2 = (3928.30^2 + 2(-3.19)(a))
a = 2418736.817ms^-2
F = ma = 556(2418735.817) = 1344817670N
Q = I suppose the work, which was mentioned above.

Is this what you got? Just wanted to try it myself. :-\
 
thats wat i got too, but i know that they're wrong becuase we have to plug our answers into this website. however, the website says the answer is incorrect...
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top