Question
If I push a 10kg box with 50.0N of force forward and the box has a coefficient of kinetic friction of .25 what is the forward acceleration of the box?
Answers
Damon
find the weight
10* 9.8 = 98 N
times friction coef = .25*98 = 24.5 N friction
net force = 50 - 24.5 = 25.5 N
F = m a
25.5 = 10 a
a = 2.55 m/s^2
10* 9.8 = 98 N
times friction coef = .25*98 = 24.5 N friction
net force = 50 - 24.5 = 25.5 N
F = m a
25.5 = 10 a
a = 2.55 m/s^2