M*g = 25kg * 9.8N./kg = 245 N. = Wt. of
the box = Normal force.
Fk = u*Fn = 0.55 * 245 = 134.75 N. = Force of kinetic friction.
a. Fp-Fk = M*a
Fp - 134.75 = M*0 = 0
Fp = 134.75 N. = Pulling force.
b. F(net) = Fp-Fk = 350 - 134.75 = 215.3
N.
c. a = (Fp-Fk)/M = 215.3/25 = 8.61 m/s^2
A 25 kg box is being pulled across a surface. The coefficient of kinetic friction between the box and the surface is 0.55.
a) if the box is moving at a constant speed, what is the pulling force?
b)If the pulling force is 350 N, what is the NET force acting on the box?
c)If the pulling force is 350 N, what is the acceleration of the box?
This is from a forces test working with newton's second law. Need the steps to figure it out.
1 answer