Physics Question (Too big to fit in this box)

Question

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)

Answer 1

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 F_k = μ_kN = μ_kmg (μ_k = 0.1) and acts on both blocks. The net force is now: F - (μ_km₁g + μ_km₂g) = 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 - μ_km₂g = m₂a.

Answer 2

Sorry, the question doesn't include data on the direction of application of the force on the block. That's a fatal flaw; it may have been diametically opposed to the direction to the second block (that's what the 2 tutors above assumed), towards the second block, tangential to the direction towards the second block, and anything in between, or even with a vertical component to it.... Gets pretty wild to solve if you're powering just one block tangentially! And, if the blocks start spinning around each other in the stationary frame of reference, of course the string tension would increase. Until it breaks. Then there might be crossover into your Language Arts composition courses: bystanders would be observed in Falling Action!....

-- Cheers, --Mr. d.

Answer 3

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)