Answer:
Usually the coefficient of friction remains unchanged
Explanation:
The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.
Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.
784 Newtons or 176.37 lbs
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
Water is, indeed, an exception. Normally when temperature drops, material shrinks. Water doesn't, water expands instead.
Answer:
P = (2 + 3) * V where V is their initial speed (total momentum)
P = 2 * 10 + 3 * Vx where Vx here would be V3
If the initial momentum is not known how can one determine the final velocity of the 3 kg obj.
Also work depends on the sum of the velocities
W (initial) = 1/2 (2 + 3) V^2 the initial kinetic energy
W (final) = 1/2 * 2 * V2^2 + 1/2 * 3 * V3^2
It appears that more information is required for this problem
