## Homework Statement

**The coreliable of kinetic friction in between the 2.0 kg block in figure and the table is 0.350.What is the acceleration of the 2.0 kg block?**

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you are assuming that the stress and anxiety is equal to the weight. That is true only if the system is not accelearting. This one is speeding up, so you have to use cost-free body diagrams of eachblock and also apply Newton"s 2nd and also 3rd legislation.

## Homejob-related Equations

F=uK*Force NormalFnet=ma## The Attempt at a Solution

I tried to take the mass"s of both the hanging and also provided them to calculate a net force on the block in the facility and also then solved for acceleration that method. Here"s my work-related.3kg*9.8=29.4N Right Side Tension1kg*9.8=9.8N Left Side Tension2kg*9.8=19.6N Force NormalF=.35*19.6N6.86N(29.4N-6.86N)-9.8N=2kg*accelerationa=6.34m/s^2and also it turns out this is wrong please aid me out give thanks to you!https://www.katifund.org/attachment.php?attachmentid=15777&d=1223407287You are watching: What is the acceleration of the 2.0 kg block?

## Homework Statement

The coefficient of kinetic friction between the 2.0 kg block in figure and the table is 0.350.What is the acceleration of the 2.0 kg block?## Homework Equations

F=uK*Force NormalFnet=ma## The Attempt at a Solution

I tried to take the mass"s of both the hanging and also offered them to calculate a net force on the block in the center and then resolved for acceleration that way. Here"s my occupational.3kg*9.8=29.4N Right Side Tension1kg*9.8=9.8N Left Side Tensionyou are assuming that the stress and anxiety is equal to the weight. That is true only if the system is not accelearting. This one is speeding up, so you have to use cost-free body diagrams of each

(29.4N-6.86N)-9.8N=2kg*accelerationa=6.34m/s^2and it transforms out this is wrong please assist me out thank you!

I"ve obtained the exact same question.I"ve got these three equations:m1*a = T1 - m1*gm2*a = T3 - T1 - μ*m2*gm3*a = m3*g - T3Are those right? Solving for a making use of the first and second equations a=(g(m3-m1))/(m3+m1) which offers a=4.9 which would certainly be the acceleration of the system if the centre mass wasn"t there.Using that worth in the net pressure equation Fnet = m3(g-a) - m1(g+a) - μ*m2*g offers Fnet=6.86. Solving Fnet=ma for a offers you a=3.43 which is not ideal the appropriate answer.Can anyone tell me wbelow I went wrong? Thanks!

I"ve obtained the very same question.I"ve got these 3 equations:m1*a = T1 - m1*gm2*a = T3 - T1 - μ*m2*gm3*a = m3*g - T3Are those right?

I"ve acquired the same question.I"ve gained these three equations:m1*a = T1 - m1*gm2*a = T3 - T1 - μ*m2*gm3*a = m3*g - T3Are those right? Solving for a making use of the first and second equations a=(g(m3-m1))/(m3+m1) which provides a=4.9 which would be the acceleration of the system if the centre mass wasn"t there.Using that worth in the net pressure equation Fnet = m3(g-a) - m1(g+a) - μ*m2*g offers Fnet=6.86. Solving Fnet=ma for a offers you a=3.43 which is not right the best answer.Can anyone tell me wright here I went wrong? Thanks!

Looks OK, although the acceleration is most likely to be in the various other direction. looks choose you cancelled out T1 and also T3, in error, using equations 1 and 3. T1 and T3 are not equal This is not true. Correct your approach of addressing the 3 equations through 3 unknowns.

Ahh you"re a life saver. Thanks a bundle!I sure am. Combine those 3 equations I posted and also you"ll obtain the appropriate answer.

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ok. and btw who is this? are you going to u of c? physics 211 or 221?. the anser i gained was 32 yet i dunt think thats appropriate. I offered the 1st and 2nd equations to acquire T1 and T3 and plugged them right into the second equation. deserve to u tell me wat i did wrong?