What is the final speed of the crate, and how do you caculate the formula?

Question

A 110-kg crate,starting from rest, is pulled across a floor with a constant horizontal force of 350 N. For the first 15 m the floor is frictionless, and for the next 15 m the coefficient of friction is 0.30. What is the final speed ofthe crete, and how do you work out the pronlem mathimatically?

Answer 1

Divide the problem into two segments...find the work done on the box along the first 15, frictionless, meters....then find the work done along the second 15 meters with friction.
The sum of the works equals the total work.
The total work done on the box equals the change in KE of the box.

W = F * d
KE = 1/2 mv^2

Since there is no friction over the first 15 meters, the net force acting on the box is the given force, 350 Newtons.
W = 350 N * 15 m = 5250 Joules

Over the second 15 meters there is a frictional force acting on the box, so the net force equals the given force minus the frictional force.
The force of friction equals the normal force * the coefficient of friction, and in this case the normal force equals the box's weight
F_f = mg * mu
where mu is the coefficient of friction
So work equals,
W = (350 N - mg*mu) * 15 m

Add these two works together,
W_1 + W_2 = W_Total

W_total = (350 N * 15 m) + (350 N - (110 kg * 9.81 m/s^2 * .30)) * 15 m
W_total = 5644.05 Joules

KE = 1/2 mv^2 = 5644.05 J
v^2 = 2 * 5644.05 J / 110 kg = 102.61 (m/s)^2
v = 10.13 m/s

Answer 2

For the first 15 m, the net force is just the applied force, and by F = ma, a = F/m = 350 N / 110 kg. Call this acceleration a_1. For the second 15 m, the net force is the applied force minus the force of friction. The friction force is equal to mgu, where u is mu, the coefficient of kinetic friction. mgu = 110 kg * 9.8 m/s^2 * 0.3. Multiply that out and subtract it from 350 N to get the net force for the second 15 m, and divide by 110 kg for the second acceleration, a_2. You now have two different accelerations for two different sections of translation. The last formula you need is V^2 = (V_0)^2 + 2ax, where V is the final velocity and V_0 is the initial velocity. x for both sections is 15 m. First apply this formula for V_0 = 0 (the initial rest condition) and a_1 to get a V that you can then use as V_0 in a second application of the formula, instead using a_2 to get a new V that is the final velocity after the second leg.

Answer 3

for the first part the plate experiences an acceleration of 350/110 m/s^2. so the initial speed is. v=(2*a*15)^(1/2). now there are two forces acting . one is 350 N and the other is 9.8*0.30. since the former is greater the crate is still being accelerated by a=(350-9.8*0.30)/110. so now use v=u^2+2as. the u is the speed calculated earlier.

Answer 4

for 1st 15 m;
a=F/m=350/110=3.1818 m/s^2

v=u'=√(u^2+2as)
=√(0+2*3.1818*15)
=9.77 m/s

for 2nd 15m;

friction, Fs= 110*9.8*0.30=323.4

∑F=ma'
=>a'=(F-Fs)/m
=(350-323.4)/110
=0.2418 m/s^2

final velocity,
v'= √(u'^2+2a's)
=√(9.77^2+2*0.2418*15)
=10.134 m/s