- #1

noob314

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## Homework Statement

A large block of mass M = 6kg is on an incline which is at an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the block and the incline is 0.3. The block is attached to a string which runs over a pulley and is connected to a smaller block of mass m = 2.4 kg.

If the block is sliding up the incline, find a) the acceleration (magnitude and direction) of the smaller block m and b) the tension in the string.

## The Attempt at a Solution

I need someone to confirm if what I'm doing is correct.

a)

[tex]\Sigma \vec{F}_{M} = Ma = T - f_{k} - Mgsin\theta[/tex]

[tex] T = M(gsin\theta + a + \mu_{k}gcos\theta) [/tex]

[tex]\Sigma\vec{F}_{m} = ma = mg - T[/tex]

[tex]= mg - M(gsin\theta + a + \mu_{k}gcos\theta)[/tex]

[tex] = ma + Ma = mg - Mgsin\theta - \mu_{k}Mgcos\theta[/tex]

[tex] = a(m+M) = mg - Mgsin\theta - \mu_{k}Mgcos\theta[/tex]

[tex] = a = \frac{mg - Mgsin\theta - \mu_{k}Mgcos\theta}{m+M}[/tex]

[tex] = a = -2.52\frac{m}{s^{2}}[/tex]

I'm guessing this is the acceleration in the y-direction, and the acceleration in the x-direction would be 0, so the magnitude is 2.52

[tex] = 2.52\frac{m}{s^{2}}[/tex] down

b)

[tex]\Sigma\vec{F}_{m} = mg - T = ma[/tex]

[tex] T = mg - ma[/tex]

[tex] = (2.4kg)(9.80 \frac{m}{s^{2}}) - (2.4kg)(2.52\frac{m}{s^{2}})[/tex]

[tex] = 17.47N[/tex]

Which looks about right, since in order for m to be moving down, mg needs to be greater than T, which in this case it is.