a 2.5kg block initially at rest on horizontal surface. A horizontal force F of magnitude 6N and a vertical force P are the applied to the block. The coefficients of friction for the block and surface are Ms=0.4 and Mk=0.25.Determine the magnitude of the frictional force acting on the block if force P is a) 8N b) 10N c) 12N

mg = 9.8 m/s^2 * 2.5kg = 24.5N

F(applied) = 6.0N
F(normal) = 24.5N - P
F(static) = Ms*(24.5 - P)
If F(static) > F(applied) = 6.0N --> F(friction) = F(applied)
If F(applied) = 6.0N > F(static) --> F(kinetic) = Mk*(24.5 - P)
a) F(normal) = 24.5N - 8.0N = 16.5N
F(static) = 0.4*16.5N = 6.6N > F(applied) = 6.0N, so F(friction) = 6.0N
b) F(normal) = 24.5N - 10.0N = 14.5N
F(static) = 0.4*14.5N = 5.8N < F(applied), so...
F(friction) = Mk*14.5 = 3.625N
c) F(normal) = 24.5N - 12.0N = 12.5N
F(static) = 0.4*12.5N = 5.0N < F(applied), so...
F(friction) = Mk*12.5 = 3.125N

In each case, there are two vertical forces: the weight and the applied force P. Multiply the sum of them by 0.4 (the static friction coefficient) to get the maximum force required to overcome friction. If this force is not exceeded by F, the block will not move. In that case, the actual friction force equals F, to prevent motion.

can you explain more

Well, well, well, let's get this block party started! Considering the horizontal force F of 6N and the vertical force P applied to our block, we need to determine the magnitude of the frictional force acting on the block. But fear not, I'm here to crunch the numbers and add a pinch of humor along the way!

a) When the force P is 8N, it seems like our block is feeling a little pushed around. In this case, let's find out the frictional force. We start by calculating the maximum static frictional force, given by Fs = Ms * N, where N is the normal force. Since the block is initially at rest, Fs = Ms * mg. Plugging in the values, we get Fs = 0.4 * 2.5kg * 9.8m/s^2. And voila! The maximum static frictional force is 9.8N (approximately).

However, since the applied force F of 6N is smaller than the maximum static frictional force, the block remains at rest, and the frictional force matches the force F: Ffriction = F = 6N.

b) Now, let's turn up the force P to 10N and see what happens. Hang on tight, things might get a little slippery! The maximum static frictional force is still the same as before, 9.8N. But this time, the applied force F is larger than the maximum static frictional force. In this case, the block starts moving, and we switch to kinetic friction.

The kinetic frictional force Fk = Mk * N, where Mk is the coefficient of kinetic friction. Plugging in the values, we get Fk = 0.25 * 2.5kg * 9.8m/s^2. After a little calculation dance, we find that Fk is approximately 6.13N. So the magnitude of the frictional force acting on the block when P is 10N is roughly 6.13N.

c) And now, our final act, let's see what happens when the force P is cranked up to 12N! Hold on to your funny bone, folks. Again, we have an applied force F larger than the maximum static frictional force. So, we're dancing with kinetic friction once more.

Using the same formula as before, Fk = Mk * N, we plug in the numbers again and find that Fk is about 6.13N. Surprise, surprise! The magnitude of the frictional force acting on the block when P is 12N is still approximately 6.13N. Looks like the block just can't catch a break!

Remember, folks, these calculations are based on the given information and the assumptions made. Now go forth, armed with the knowledge of friction forces and enjoy the amusing world of physics! Keep rolling!

To determine the magnitude of the frictional force acting on the block, we need to consider the different cases where P is 8N, 10N, and 12N. We'll start by calculating the normal force and then use the coefficients of friction to determine the frictional force.

1. Calculate the normal force (N):
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the vertical force P partially balances out the weight of the block and affects the normal force. The formula for the normal force is N = mg - P, where m is the mass of the block.

Given:
Mass of the block (m) = 2.5 kg
Gravity (g) = 9.8 m/s^2

a) When P = 8N:
N = (2.5 kg)(9.8 m/s^2) - 8 N
N = 24.5 N - 8 N
N = 16.5 N

b) When P = 10N:
N = (2.5 kg)(9.8 m/s^2) - 10 N
N = 24.5 N - 10 N
N = 14.5 N

c) When P = 12N:
N = (2.5 kg)(9.8 m/s^2) - 12 N
N = 24.5 N - 12 N
N = 12.5 N

Now that we have determined the normal force for each case, we can proceed to find the frictional force.

2. Calculate the frictional force (f):
The frictional force depends on whether the block is at rest or in motion. There are two types of friction: static friction (when the block is at rest) and kinetic friction (when the block is in motion). We need to consider both cases.

a) When P = 8N:

i) Determine if the block is at rest or in motion:
Since the block is initially at rest, we need to calculate the maximum static friction.
The formula for the maximum static friction (fs max) is fs max = Ms * N.

Given:
Coefficient of static friction (Ms) = 0.4
Normal force (N) = 16.5 N

fs max = (0.4)(16.5 N)
fs max = 6.6 N

The frictional force (f) acting on the block when it's not in motion is equal to the maximum static friction (fs max).

f = 6.6 N

b) When P = 10N:

i) Determine if the block is at rest or in motion:
Since the block is initially at rest, we need to calculate the maximum static friction.
fs max = (0.4)(14.5 N)
fs max = 5.8 N

f = 5.8 N

c) When P = 12N:

i) Determine if the block is at rest or in motion:
Since the block is initially at rest, we need to calculate the maximum static friction.
fs max = (0.4)(12.5 N)
fs max = 5 N

f = 5 N

To summarize, the magnitude of the frictional force acting on the block is:
a) When P = 8N: 6.6 N
b) When P = 10N: 5.8 N
c) When P = 12N: 5 N

(a) If P = 8N, the maximum static friction force is Ms*(M*g + P) =

(0.4)(24.5 + 8) = 13 N
Since the applied force is less than that, no motion will take place and the actual friction force will equal the applied horizontal force, 6N .