Wow ! This is not simple, Shoot, and I give you a lot of credit
and an extra merit badge if you're generally keeping up with it.
I scratched my head for a few minutes, and I think I've got it.
Here's what I think is going on:
KE₁ = KE of the box before pushing
(1/2) (m) (speed²) = 10 x 2² = 40 joules
KE₂ = KE of the box after pushing 3m
(1/2) (m) (speed²) = 10 x 4² = 160 joules
The box gained (160 - 40) = 120 J of kinetic energy.
Now look at the cluttered force diagram.
Cat's component of force in the direction of motion is 120N.
That's the part of her force that does the work on the box.
How much work does she do ?
(force) x (distance) = (120N) x (3m) = 360 joules .
Only 120 J of that energy showed up as increased kinetic energy
of the box. The other 240J of her hard-earned work was consumed
by friction.
Work of friction = (Friction force) x (distance)
240 J = (friction force) x (3 m)
240 J / 3 m = friction 'force' = 80 N .
I think that's it.
What I did was:
-- Find the work that Cat did.
-- Find the increase in the kinetic energy of the box.
-- The difference ... the 'missing energy' ... was the work done
by friction in the same distance.
Does this do anything for you ?
<h3>
Answer: Approximately 4.67 m/s^2</h3>
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Explanation:
Let's say you want to push the brick to the right. The free body diagram will have an arrow pointing right on the rectangle (the brick) and the arrow is labeled with 35 N.
Friction always counteracts whatever force you apply. The friction force arrow will point left and be labeled with 7 N.
The net horizontal force is therefore 35-7 = 28 N and the direction is to the right. The positive net force means you've overcome the force of friction and the brick is moving.
F = 28 is the net force
m = 6 is the mass
a = unknown acceleration
F = m*a .... newton's second law
28 = 6a
6a = 28
a = 28/6
a = 4.67
The acceleration of the brick is approximately 4.67 m/s^2
This means that for every second, the brick's velocity is increasing by about 4.67 m/s.
