Question
5.A crate with a mass of 100kg accelerates with a magnitude of 2.0 m/s2 when it is pushed with a 250N.
a.What is magnitude of the force of friction opposing this motion?
b.What is the normal force of the crate?
c.Using a and b, find the coefficient of friction.
d.If the crate starts from rest, find the distance the crate travels in 3s.
a.What is magnitude of the force of friction opposing this motion?
b.What is the normal force of the crate?
c.Using a and b, find the coefficient of friction.
d.If the crate starts from rest, find the distance the crate travels in 3s.
Answers
Henry
a. Fn = ma = 100 * 2 = 200N = Net force.
Fn=Fap - Ff=200, Fap = Force applied.
250 - Ff = 200,
-Ff = 200 - 250 = -50,
Ff = 50N = Force of friction.
b. Fc = mg = 100kg * 9.8N/kg = 980N =
Force of the crate = Normal force.
c. u = Ff/Fc 50/980 = 0.05 = coefficient of friction.
d. d = Vo*t + 0.5at^2,
d = 0 + 0.5*2*3^2 = 9m.
Fn=Fap - Ff=200, Fap = Force applied.
250 - Ff = 200,
-Ff = 200 - 250 = -50,
Ff = 50N = Force of friction.
b. Fc = mg = 100kg * 9.8N/kg = 980N =
Force of the crate = Normal force.
c. u = Ff/Fc 50/980 = 0.05 = coefficient of friction.
d. d = Vo*t + 0.5at^2,
d = 0 + 0.5*2*3^2 = 9m.