Work and Energy Problems [help]

AI Thread Summary
The discussion focuses on calculating work done in two scenarios: accelerating an electron and stopping a car. To find the work required to accelerate an electron from rest to a speed of 5.0 x 10^6 m/s, the kinetic energy formula KE = 1/2 mv^2 is applicable. For the car, a 1000-kg vehicle traveling at 100 km/h requires 385,809 Joules of work to stop, calculated using the same kinetic energy formula. Participants emphasize the importance of unit conversion and significant figures in calculations. Overall, the thread highlights the application of kinetic energy principles in solving work and energy problems.
krypt0nite
Messages
31
Reaction score
0
1) How much work does it take to accelerate an electron(mass=9.11x10^-31kg)
from rest to a speed of 5.0x10^6 ms?

I know you have to use more than 1 equation to solve it but I can't seem to get close to an answer.

2) How much work must be done to stop a 1000-kg car traveling at 100km/h?

Is it me or does this question not have enough info.
 
Physics news on Phys.org
krypt0nite said:
Is it me or does this question not have enough info.
It's you. :smile: Remember that work done on an object equals its change in KE.
 
krypt0nite said:
2) How much work must be done to stop a 1000-kg car traveling at 100km/h?

KE = 1/2mv^2
KE = 1/2(1000)(27.278)^2
KE = 385809 Joules
So the work to stop it would be that much?
 
And I assume the 1st question is done the same way?
 
krypt0nite said:
KE = 1/2mv^2
KE = 1/2(1000)(27.278)^2
KE = 385809 Joules
So the work to stop it would be that much?
That's the idea. (Check your conversion of 100km/hour.) (Don't state so many significant figures in your answer.)

And, yes, the first problem is the same idea.
 
Doc Al said:
That's the idea. (Check your conversion of 100km/hour.) (Don't state so many significant figures in your answer.)

And, yes, the first problem is the same idea.
Thanks SOO MUCH!
 
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