Answer:
1. increases
2. increases
3. increases
Explanation:
Part 1:
First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:
F1 - fs = 0.
And this friction force fs is:
fs = Nμs,
where μs is the static coefficient of friction, and N is the normal force.
Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:
N = mg + F2.
So, F2 is increasing, that means fs is increasing too.
Part 2:
As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.
Part 3:
In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.
I would expect it to be slightly basic.
Answer:
<em>The distance is now 4d</em>
Explanation:
<u>Mechanical Force</u>
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
F = m.a
Where a is the acceleration of the object.
The acceleration can be calculated by solving for a:

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

If the block starts from rest, vo=0:

Substituting the value of the acceleration:

Simplifying:

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

And the distance is now:

Dividing d'/d:

Simplifying:

Thus:
d' = 4d
The distance is now 4d
Answer:
The answer to your question is given below.
Explanation:
Mechanical advantage (MA) = Load (L)/Effort (E)
MA = L/E
Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)
VR = l/e
Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100
Efficiency = Wd/Wp x100
Recall:
Work = Force x distance
Therefore,
Work done by machine (wd) = load (L) x distance (l)
Wd = L x l
Work put into the machine (Wp) = effort (E) x distance (e)
Wp = E x e
Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)
Efficiency = Wd/Wp x100
Efficiency = (L x l) / (E x e) x 100
Rearrange
Efficiency = L/E ÷ l/e x 100
But:
MA = L/E
VR = l/e
Therefore,
Efficiency = L/E ÷ l/e x 100
Efficiency = MA ÷ VR x 100
Efficiency = MA / VR x 100
<h3>Answer</h3>
m/s^2 (meter per sec square)
Explanation:
acc = change in velocity/time
= distance/time
----------------
time
= m/s
------
s
=m/s^2