Two boxes of fruit on a frictionless horizontal surface are connected by a light string as in the figure below, where m1 = 16 kg and m2 = 28 kg. A force of F = 76 N is applied to the 28-kg block.

https://www.webassign.net/sercp9/4-p-069-alt.gif

(a) Determine the acceleration of each block and the tension in the string.

1.) acceleration of m1 __?__m/s2

2.) acceleration of m2 __?__m/s2

3.) tension in the string __?__N

(b) Repeat the problem for the case where the coefficient of kinetic friction between each block and the surface is 0.10.

1.) acceleration of m1 __?__m/s2

2.) acceleration of m2 __?__m/s2

3.) tension in the string __?__N

### 1 Answer

a) Both accelerate at the same rate.

Acceleration = (force/ mass) = 76/(16 + 28) = 1.73m/sec^2.

Tension in connecting string = (mass x acceleration) = (16 x 1.73) = 27.68N.

b) Normal force = (16 + 28) x g = 44 x 9.8 = 431.2N.

Friction force = (431.2 x 0.1) = 43.12N. total.

Net accelerating force = (76 – 43.12) = 32.88N.

Both will still accelerate at the same rate.

Acceleration = (force/ mass) = 32.88/(28 + 16) = 0.75m/sec^2.

Normal force of m1 alone = (16 x g) = 156.8N.

Friction component of m1 alone = (156.8 x 0.1) = 15.68N.

Tension due to acceleration alone, no friction = (mass x acceleration) = 16 x 0.75 = 12N.

Add to that the friction force, = total tension of (12 + 15.68) = 27.68N. in the string.