Haylee Bradtke
Dec 21, 2024
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?
Two boxes of fruit are placed on a frictionless horizontal surface and connected by a light string, as shown in the figure (link: https://www.webassign.net/sercp9/4-p-069-alt.gif). The masses of the boxes are m1 = 16 kg and m2 = 28 kg. A force of F = 76 N is applied to the 28-kg box.
(a) Determine the following:
1. The acceleration of the 16-kg box (m1): ? m/s²
2. The acceleration of the 28-kg box (m2): ? m/s²
3. The tension in the string: ? N
(b) Repeat the calculations for the case where the coefficient of kinetic friction between each box and the surface is 0.10.
1. The acceleration of the 16-kg box (m1): ? m/s²
2. The acceleration of the 28-kg box (m2): ? m/s²
3. The tension in the string: ? N
1 Answers
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.
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