Finding Ek with force, mass, and time.

  • Thread starter Thread starter rowkem
  • Start date Start date
  • Tags Tags
    Force Mass Time
AI Thread Summary
A 10kg box is pushed on a frictionless surface with a 10N force, resulting in an acceleration of 1 m/s². After 10 seconds, the final velocity is calculated to be 10 m/s. Using the kinetic energy formula, the final kinetic energy is determined to be 500J. The calculation is confirmed by considering the work done by the force over the distance traveled. The conclusion is that the calculated kinetic energy of 500J is indeed correct.
rowkem
Messages
48
Reaction score
0

Homework Statement



A 10kg box at rest is being pushed on a frictionless surface by a force of 10N. The force is in the same direction as the motion of the box. 10 seconds later, how much kinetic energy will this box have?

Homework Equations



1. F=ma
2. Vf=Vi+at
3. Ek=0.5mv2

The Attempt at a Solution



I used formula 1 along with the F and mass to find the acceleration to be 1 m/s2. Then, I plugged that into formula 2, given an initial velocity of 0, acceleration of 1 m/s2, and a time of 10s. This gave me Vf at 10m/s. I used that in formula 3, with mass being 10kg. This gave me a final Ek of 500J.

Is this correct? 500J seems a little high but, I've no clue how else to do this. Thanks in advance.
 
Physics news on Phys.org
1J is the energy exerted by a force of 1N thaat moves an object by 1m. The force in this case is 10N and the distance is 50m(0.5at^2=100/2), so 500J seems fine.
 
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