Calculate Distance Travelled in 3s with 12.5N Force on 13.5kg Cart

  • Thread starter Thread starter Pencil
  • Start date Start date
  • Tags Tags
    Force
AI Thread Summary
To calculate the distance traveled by a 13.5 kg shopping cart pushed with a force of 12.5 N over 3 seconds, the acceleration is determined using F = ma, resulting in 0.93 m/s². The final velocity after 3 seconds is calculated to be 2.79 m/s. The distance traveled can be found using the equation x = (Vf²)/(2a), which corrects the earlier manipulation error. The correct calculation yields a distance of approximately 4.0 meters. Rounding errors should be avoided for accurate results.
Pencil
Messages
17
Reaction score
0
In a grocery store, you push a 13.5 kg shopping cart with a force of 12.5 N. If the cart starts at rest, how far does it move in 3.00 s?

Here's what I did:
m=13.5 kg
f=12.5 N
t=3s
v(i)=0
F=ma
a=f/m
12.5/13.5=0.93m/s^2=accel.
V(final)=0+at= 0+(.93)(3)=2.79
Vf^2=Vi^2+2a(deltax)= solve for change in X= (2a)Vf^2 (Vf = final velocity, Vi= initial velocity)
so, (2)(.93)*(2.79)^2= 14.5 m.
Ok, so what did I do wrong? I can't figure it out...
 
Physics news on Phys.org
Pencil said:
Vf^2=Vi^2+2a(deltax)= solve for change in X= (2a)Vf^2 (Vf = final velocity, Vi= initial velocity)
Here's your problem, you've just made a mistake on your manipulation;

v_{f}^{2} = v_{i}^{2} + 2ax \Rightarrow x = \frac{v_{f}^{2}}{2a}

Also, becareful of inducing rounding errors.
 
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