Question
A force of 1000 N is applied to a 1200 kg car. If the coefficient of friction is 0.04, what is the car's acceleration?
Answers
drwls
a = Fnet/m
The net force is
Fnet = 1000 - M*g*Uk
= 1000 - 470.4 = 529.6 N
a = 529.6/1200 = 0.441 m/s^2
The net force is
Fnet = 1000 - M*g*Uk
= 1000 - 470.4 = 529.6 N
a = 529.6/1200 = 0.441 m/s^2
Maria Ellah
thank you!
jocynt
how did you cope in this answer what is the formula and solution?
Shraddhanjali
Help full👍