- #1
Sciencelover91
- 11
- 1
Homework Statement
A wooden pallet carrying a load of 600 kg rests on a wooden floor.
(s= .28 and k= .17)
a. A forklift driver decides to push it horizontally instead of lifting it. What force must be
applied to just get the pallet moving from rest?
b. After a bit of time, the driver pushes the pallet and it slides. How fast is the pallet
moving after 0.5 seconds of sliding under the same force you calculated in part a?
Draw a free body diagram – don’t forget about friction!
c. If the forklift driver stops pushing at 0.5 seconds, how far does the pallet slide before
coming to a stop?
Homework Equations
Fs= us (Fn)
Fk= uk (Fn)
F= ma
V^2 = vo^2 + 2 a (change in X)
V=vo+ at
The Attempt at a Solution
A) I know that to just get the pallet moving from rest means right when it moves so it's not moving, making it static friction so I did: Fs = .28 (600kg X 9.81 m/s^2) = 1648 N
B) I did f= ma because the problem said under the same force from part A so 1648 N = 600 kg (a) and got a = 2.75 m/s^2
Then I did v = vo + at so it becomes v = 2.75 m/s^2 (.5 s) = 1.37 m/s^2. However my answer is wrong because I didn't use Fk= uk as it is now kinetic friction. The correct answer is supposed to use Fs - Fk= ma and is a = 1.07 m/s^2 and v = .54 m/s but I don't know why. Why do you subtract the two frictions and what happened to using the Applied force?
C) I'm using the correct velocity (.54 m/s) and plug it into V^2 = vo^2 + 2 a (change in X) but I don't get the right answer which is .0875 m. I did (.54 m/s)^2 = 0 + 2 (1.07m/s^2)(change in x) and got change in x = .136 m. I do not know how to get the correct answer if it's the right equation.
Last edited: