Question

Q. A wooden pallet carrying a load of 600 kg rests on a wooden floor.
A forklift driver decides to push it without lifting it. What force must be applied to just get the pallet
moving?
After a bit of time, the pallet begins to slide. How fast is the pallet moving after sliding under the same
force you calculated in part a for half a second?
If the forklift stops pushing, how far does the pallet slide before coming to a stop?​

Answer 1

Answer:

To get the pallet started, the driver must push it with a force equal to the maximum static friction. Once the pallet starts moving, the coefficient of friction drops from its static value to its kinetic value. But the forklift is still pushing with 1650 N of force.

Explanation:

Four forces are acting on the pallet: the downward pull of Earth's gravity, the normal force of the floor pushing up, the forward push of the forklift, and the backward resistance of friction. Weight and normal are equal throughout this example since the floor is level. Friction changes from static to kinetic — static friction initially since the pallet isn't moving initially, then kinetic friction once the pallet gets going. The push also changes from nothing to the value needed to get the pallet moving, then back to nothing after 0.5 seconds of motion.

Answer 2

Total 1646N force should be applied to get the pallet moving.

The pallet will move with velocity 0.54m/s.

Pallet will come 0.087m far before coming to stop.

Here we can figure out that total four forces are acting on the pallet which are Gravitational pull downwards, normal force upwards, front force of forklift and the backward frictional force. Here weight and normal are equal but frictional force tends to change from static to kinetic. Static force is at initial rest position and will change to kinetic when it would start moving.

P = fs = μsmg

P = (0.28)(600)(9.8)

P = 1646 N

As the pallet starts moving coefficient of friction changes from static to kinetic.

fk = μkN = μkmg

fk = (0.17)(600)(9.8)

fk = 1000N

∑F = P - fk

∑F = 1646N - 1000N

∑F = 646 N.

a = $\frac{∑F}{m}$

$a = \frac{646}{600}$

a = 1.08$\frac{m}{s^{2} }$

$v = v_{0} + at\\ \\v = \frac{1.08}{0.5}\\ \\v = 0.54\frac{m}{s}$

As the forklift stops to push, kinetic friction is the net force now. Also this causes acceleration opposite the direction of motion.

$\frac{∑F}{m} = \frac{fk}{m} \\ \\a = \frac{-1000}{600}\\ \\a = -1.67\frac{m}{s^{2} }$

$v^{2} = v_{0} ^{2} + 2as\\\\ s = -(0.54)(22)(-1,67) = 0.087m$