Physics question?

Question

The coefficient of static friction between a block and a horizontal floor is 0.40, while the coefficient of kinetic friction is 0.15. The mass of the block is 5.0 kg. A horizontal force is applied to the block and slowly increased. (a) What is the value of the applied horizontal force at the instant that the block starts to slide? (b) What is the net force on the block after it starts to slide?

Answer 1

1+1=2

Answer 2

You're using a model which states that the force of static friction is less than or equal to the coefficient times the normal force. Since the floor is horizontal, the normal force will equal the weight of the block. (Do you understand why?) The equlibrium point will be where the force applied to the block is exactly equal to the maximum static-friction force, so use the coefficient times the weight of the block for part a.

For b, realize that the force of kinetic friction is also the coefficient of friction times the normal force, but the coefficient of kinetic friction is significantly less than the coefficient of static friction. The block will start to slide when the force equals the force you calculated in (a) above - just take the difference between that force and the new force of friction and you'll have the net force.

Answer 3

a)
Just as the block is about to slide, the static friction is still at play.

The force required to get the block to that point is equivalent to the coefficient of static friction times the normal force.

Force applied = u(s) * N
=u(s) * g * mass
= 0.40 * 9.8 (m/s^2) * 5.0 kg
=19.6 (kg*m/s^2) or 19.6 Newtons

b)
The net force acting on the block just as it is sliding involves kinetic friction

u(k)*N = 0.15 * 9.8 (m/s^2) * 5 kg
=7.35 Newtons

Answer 4

The static force is affected by the weight. Since the component of the forces are at 90 degrees to each other, the question becomes very much simpler.

The force required to overcome static friction is 0.4 x 5 x g (9.81)
= 19.62 or 20N (if you assume your g to be 10).
Once the block is moving, the force is reduced to 0.15 x 5 x g (9.81) = 7.3575 or 7.5N (approx).

Answer 5

When your force reaches (.4)*(9.81m/s^2)(5kg) the net force = 0
i.e. your force is pusing 19.62N on the block and the frinction force is pushing the equal force in the opposite direction. So as soon as the F>19.62 the block starts to move. Now as described the force is slowly increased but I will assume that it is stopped once the block begins to move so when F=19.620000001 or whatever we'll just simplify to 19.62 the friction force now is equal to .15*9.81*5=7.3575 so the net force = 19.62-7.3575 = 12.2625 N. However if the force continues to increase after the block starts moving then so will the net force as a function of:
f(netforce)=f(F)-7.3575

Answer 6

I took Physics back in high school in 1960. Frankly, I don't remember how to do this problem. And, I can tell you in the past 46 years I have not had any opportunity to think about this problem. However, "if" you are going to be some sort of space scientist or mechanical engineer then you should have your text books readily available so that you can keep current on how to solve these problems. Good luck!

Answer 7

F=ma. force equals mass x acceleration. Calculate the acceleration as m/s^2. You can plug in mass and acceleration. Figure up the force from there. F = mg also. where mass and gravity are known. calculate from there.

Answer 8

wt X FC = Force will help

Answer 9

Honestly, this is why I quit physics and went to math.

Answer 10

umm, I dont know, sorry