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Physics Question (Too big to fit in this box)

Two blocks on a frictionless horizontal surface are connected by a light string, where m1 = 11 kg and m2 = 22 kg. A force of 80 N is applied to the 22 kg block.

(a) Determine the acceleration of each block and the tension in the string.
m/s2 (acceleration of m1)
m/s2 (acceleration of m2)
N (tension in the string)
(b) Repeat the problem for the case where a coefficient of kinetic friction between each block and the surface is 0.1.
m/s2 (acceleration of m1)
m/s2 (acceleration of m2)
N (tension in the string)

Daniel O. | Math and Physics Tutor, with a math and physics degreeMath and Physics Tutor, with a math and ...
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A free body diagram can be helpful here, to find the direction of all the forces acting on each block.

The acceleration of each block will be equal, since they're connected by the string, so the 80N force acts to accelerate the total mass (m1+m2). The tension does not effect the acceleration of the total mass, it acts equally at both ends of the rope, so it balances out - it acts to slow the acceleration of the 22kg block, and causes of the acceleration of the 11kg block (again, easy to see with a free body diagram).

Part (a): To find the acceleration, apply Newton's second law, F = ma where m is the total mass (m1+m2) and F is the 80N force, as that's the net force acting on the total mass.

Once you find the acceleration, you can use it to find the tension, by T = m2a since the only horizontal force acting on m2 is the tension (if it's not clear, we're still using F = ma, but in this case F = T). We can also find the tension from m1, but it's easier with m2.

Part (b), we use the same process but have to account for friction, which acts to oppose acceleration. Friction force is given by Fk = μkN = μkmg (μk = 0.1) and acts on both blocks. The net force is now: F - (μkm1g + μkm2g) = ma, where F = 80N and m is the total mass.

Once you find the acceleration for part (b), we can find the tension as above, but this time we have both tension and friction acting in opposite directions on m2, so the net force is: T - μkm2g = m2a.

Thanks you

I'm having trouble still finding the tension in part a and part b.

Once you find the acceleration, you can substitute that into the equation for tension, along with the values for m2, g, friction, and then solve for T.

Robert J. | Certified High School AP Calculus and Physics TeacherCertified High School AP Calculus and Ph...
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a)Draw free body diagram, from which you can write two equations based on Newton's second law:

80-T = m2a ......(1)

T = m1a ......(2)

(1)+(2) and simplify, a = 80/(m1+m2)

Therefore, T = 80m1/(m1+m2)

b) Similar to a). Adding friction forces, you have

80 - T - 0.1m2g = m2a ......(1)

T -0.1m1g = m1a ......(2)

(1)+(2) and simplify, a = 80/(m1+m2) - 0.1g

T = m1(a+0.1g) = 80m1/(m1+m2)