Solving Friction Problems: Calculating Work

[needhelp83]

1)If the coefficient of kinetic friction between a 35-kg crate and the floor is 0.3, what horizontal force is required to move the crate at a steady speed across the floor?

FN=mg=(35 kg)(9.8 m/s2)=343 N
Ffr= kF N=(0.30)(343 N)=102.9 N




2)A 1200-N crate rests on the floor. How much work is required to move it at constant speed
a) 4 m along the floor against a friction force of 230 N
b) 4 m vertically?

a)
W G = mgxcos90°=0
W N = FNxcos90°=0
W net = (Fnet)xx=( Fnet – Ffr)x= (1200 N – 230 N)(4 m)= 3880 J

b)
W G = mgxcos90°=0
W N = FNxcos90°=0
W net = (Fnet)x=( Fnet – Ffr)x= (1200 N – 0)(4 m)= 4800 J

Just wanting to make sure I am doing everything correctly.

 

[berkeman]

-1- looks good

-2a- They give you the force that is opposing your push. Work is force x distance.

-2b- The force opposing the lift is just the weight, so you have the correct answer. But I don't really understand what you are trying to say with the two equations =0 part. The work done is just the change in potential energy for the object between the two different heights.

 

[needhelp83]

2a)

Would it just be
W=Fd
W=230N*4=920 J

 

[berkeman]

2a)

Would it just be
W=Fd
W=230N*4=920 J

That would be my answer. How'd we do?