(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider the following diagram:

Assume the strings and pulleys have negligible masses and the coefficient of kinetic friction between the 2.0 kg block and the table ([itex]\mu[/itex][itex]_{k}[/itex]) is .12. What is the acceleration of the 2.0 kg block?

2. Relevant equations

Net force = T[itex]_{2}[/itex] - (T[itex]_{1}[/itex] + [itex]\mu[/itex][itex]_{k}[/itex]F[itex]_{g}[/itex]), where T[itex]_{2}[/itex] is the tension from the 3 kg block, T[itex]_{1}[/itex] is the tension from the 1 kg block, and F[itex]_{g}[/itex] is the force of gravity on the 2 kg block.

a= Net force/mass

3. The attempt at a solution

This seems like a really straightforward problem, but none of the multiple choice answers match my result so I'm confused as to what exactly I'm missing.

I drew a free-body diagram for the 2 kg block with the normal force pointing up, the force of gravity pointing down, T[itex]_{1}[/itex] + [itex]\mu[/itex][itex]_{k}[/itex]*F[itex]_{g}[/itex] pointing left (presumably the block will accelerate to the right, so the friction force points in the opposite direction), and T[itex]_{2}[/itex] pointing right.

T[itex]_{1}[/itex] = 1 kg * 9.8 m/s^2 = 9.8 N.

[itex]\mu[/itex][itex]_{k}[/itex]*F[itex]_{g}[/itex] = .12 * 2 * 9.8 m/s^2 = 2.4 N

T[itex]_{2}[/itex] = 3 kg * 9.8m/s^2 = 29.4 N.

Net force = 29.4 N - (9.8+2.4) = 17.2 N

a = 17.2 N/ 2 kg = 8.6 m/s^2.

All of the options for answers are below 4.6 m/s^2.

I'm not looking for the answer; I'd like to just find out conceptually what I'm missing so I can solve it for myself.

What am I missing?

Thanks in advance,

m3rc

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# Acceleration due to Tension

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