A mass (m1) is connected by a light string that passes over a massless frictionless pulley to a mass (m2) sliding on a horizontal surface. The coefficient of kinetic friction between the m2 and the surface is 0.25. If m1=1.0kg and m2=2.0kg then what is the acceleration of the system?

I know I need to use the equation:
a= [(m1g) - (μk)(m2g)]/(m1-m2).
The answer for my acceleration is supposed to equal 1.6

*I'm having trouble identifying/obtaining the value for μk.*
I know Fk=μk(Fn), and Fn=(m2)(g)=19.6

I tried taking the answer for acceleration and solving for the μk value I'm supposed to have and I got 0.41836735, but I have no idea how I would get that number to solve for the acceleration to begin with.

mk is given as .25, the kinetic force of friction constant.

Shouldn't the denominator be m1+m2? both are accelerating.

Net force=total mass*a

total mass = m1+m2 = 3

down force on m1 = 9.8 (m1.g)
fric force on m2 = 2*9.8/4 = 9.8/2
(m2.g.mk, cuz mk = 0.25 = 1/4)

net force = 9.8 - 9.8/2 = 9.8/2

a = net force/total mass = 9.8/2/3 = 1.6

:)

To solve for the kinetic friction coefficient (μk), you can use the following steps:

1. Begin with the equation for the acceleration of the system:

a = [(m1g) - (μk)(m2g)] / (m1 + m2)

2. Substitute the given values:

m1 = 1.0 kg
m2 = 2.0 kg

3. To find μk, we need to identify the normal force (Fn) acting on m2. The normal force is equal to the weight of m2, which is m2g:

Fn = (m2)(g)
Fn = (2.0 kg)(9.8 m/s^2)
Fn = 19.6 N

4. Next, we can rewrite the equation for acceleration by substituting the values of m1, m2, and Fn:

a = [(1.0 kg)(9.8 m/s^2) - (μk)(2.0 kg)(9.8 m/s^2)] / (1.0 kg + 2.0 kg)

5. Simplify the equation:

a = (9.8 N - 19.6 μk N) / 3.0 kg
a = 9.8 N / 3.0 kg - (19.6 N / 3.0 kg)(μk)

6. Now, equate this equation to the given acceleration value, a = 1.6 m/s^2, and solve for μk:

1.6 m/s^2 = 9.8 N / 3.0 kg - (19.6 N / 3.0 kg)(μk)

Multiply both sides of the equation by 3.0 kg to eliminate the denominator:

(1.6 m/s^2)(3.0 kg) = 9.8 N - (19.6 N)(μk)

4.8 kg m/s^2 = 9.8 N - 19.6 N μk

Rearrange the equation to solve for μk:

19.6 N μk = 9.8 N - 4.8 kg m/s^2

μk = (9.8 N - 4.8 kg m/s^2) / 19.6 N

μk ≈ 0.2561

Now that you have obtained the value for μk, you can substitute it back into the original equation for acceleration and solve for the acceleration of the system:

a = [(m1g) - (μk)(m2g)] / (m1 + m2)
a = [(1.0 kg)(9.8 m/s^2) - (0.2561)(2.0 kg)(9.8 m/s^2)] / (1.0 kg + 2.0 kg)
a ≈ 1.6 m/s^2

Therefore, the acceleration of the system is approximately 1.6 m/s^2, as expected.

To solve for the acceleration of the system, you correctly identified the equation to use:

a = [(m1g) - (μk)(m2g)] / (m1 + m2)

The value you are missing is the coefficient of kinetic friction (μk). In order to find μk, we need to use the equation Fk = μkFn, where Fk is the force of kinetic friction and Fn is the normal force.

The normal force, Fn, in this situation is equal to the weight of the sliding mass (m2) which is given by:

Fn = m2g

Given that m2 = 2.0 kg and g = 9.8 m/s², we can find the normal force:

Fn = (2.0 kg)(9.8 m/s²) = 19.6 N

Now, we need to find the force of kinetic friction, Fk. The force of kinetic friction is given by:

Fk = μkFn

Substituting the known values:

Fk = μk(19.6 N)

The coefficient of kinetic friction (μk) is 0.25 (given in the problem).

Now, substituting this value and the other given values into the equation for acceleration:

a = [(m1g) - (μk)(m2g)] / (m1 + m2)
= [(1.0 kg)(9.8 m/s²) - (0.25)(2.0 kg)(9.8 m/s²)] / (1.0 kg + 2.0 kg)
= [(9.8 - 4.9) N] / 3.0 kg
= 4.9 N / 3.0 kg
= 1.63333... m/s²

Rounding to the correct number of significant figures, the acceleration is approximately 1.6 m/s².

So the correct value for acceleration is indeed 1.6, and solving for μk will not lead to that value. Make sure to use the given coefficient of kinetic friction (μk = 0.25) in the calculation to obtain the correct result.