What Is the Coefficient of Friction for a Toboggan with Added Weight?

[xdeanna]

A 12kg toboggan is pulled along at a constant velocity on a horizontal surface by a horizontal force of 11N.
a) What is the force of gravity on the toboggan?
b) What is the coefficient of friction?
c) How much horizontal force is needed to pull the toboggan at a constant velocity if two 57kg girls are sitting in it?

m=12kg
F_A= 11N

a) F_g=F_N
mg=F_N
12(9.8)=F_N
118N=F_N=F_g

F_g=118N

b) F_A=F_f; constant vel.
F_K=F_A=11N

Fk=μ_KF_N
11N= μ_K(118N)
μ_K=11N/118N
μ_K=0.093

c)m= 12kg + 2(57kg)= 126kg

F_g=F_N
mg=F_N
126(9.8)=F_N
1234.8N=F_N


m=126kg
F_N=1234.8N
F_K=F_A=?
im stuck here..

 

[Ahalp]

A 12kg toboggan is pulled along at a constant velocity on a horizontal surface by a horizontal force of 11N.
a) What is the force of gravity on the toboggan?
b) What is the coefficient of friction?
c) How much horizontal force is needed to pull the toboggan at a constant velocity if two 57kg girls are sitting in it?

m=12kg
F_A= 11N

a) F_g=F_N
mg=F_N
12(9.8)=F_N
118N=F_N=F_g

F_g=118N

b) F_A=F_f; constant vel.
F_K=F_A=11N

Fk=μ_KF_N
11N= μ_K(118N)
μ_K=11N/118N
μ_K=0.093

c)m= 12kg + 2(57kg)= 126kg

F_g=F_N
mg=F_N
126(9.8)=F_N
1234.8N=F_N


m=126kg
F_N=1234.8N
F_K=F_A=?
im stuck here..

You have the coefficient of friction from part b. You can use this value to find the force of kinetic friction acting on the toboggan+kids.

And because the tobaggon must move at a constant velocity, the net force must be zero. Thus, like you wrote, Fa = Fk.

 

[xdeanna]

You have the coefficient of friction from part b. You can use this value to find the force of kinetic friction acting on the toboggan+kids.

And because the tobaggon must move at a constant velocity, the net force must be zero. Thus, like you wrote, Fa = Fk.

omg i feel stupid now :|
thanks lol