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Aliun [14]
3 years ago
6

How does the force of friction change when the normal force changes?

Physics
1 answer:
77julia77 [94]3 years ago
8 0
The friction force is given by:

F_f= \mu F_N

Where μ is the coefficient of friction (static or kinetic) and F_N is the normal force.  So we say the friction force is directly proportional to the normal force with a constant of proportionality equal to the coefficient of friction.
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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22
alexgriva [62]
1) The law of motion of the projectile is
h(t) = 2+22.5 t-4.9 t^2
To find the velocity, we should compute the derivative of h(t):
v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t
So now we can calculate the speed at t=2 s and t=4 s:
v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s
v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
v(t)=0
So we have
22.5-9.8 t=0
And solving this we find
t=2.30 s

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
2+22.5t -4.9 t^2 = 0
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
t=4.68 s
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36  m/s
with negative sign, because it is directed downward.
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