- #1
Trista
- 33
- 0
So, there are two boxes of fruit on a frictionless horizontal surface and are connected by a light string. m1=10kg, m2=20kg. A force of 50N is applied to the 20-kg box.
There are two questions to answer, I already answered a:
Determine the acceleration of each box, and the tension in the string:
a= F/m1+m2 = 50N/30kg = 1.7 m/s^2 and T=m1a ... T=10kg(1.7 m/s) = 17 N.
This is the one I can't figure out, b:
Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.10. So, mu k = .10
I figured that the equation to use is Newtons 2nd Law, sum Fx=T-fk=m1a1. But I can't get it to work. I actually know the answer, but I don't know how to get there. Any help?
There are two questions to answer, I already answered a:
Determine the acceleration of each box, and the tension in the string:
a= F/m1+m2 = 50N/30kg = 1.7 m/s^2 and T=m1a ... T=10kg(1.7 m/s) = 17 N.
This is the one I can't figure out, b:
Repeat the problem for the case where the coefficient of kinetic friction between each box and the surface is 0.10. So, mu k = .10
I figured that the equation to use is Newtons 2nd Law, sum Fx=T-fk=m1a1. But I can't get it to work. I actually know the answer, but I don't know how to get there. Any help?