Assume the three blocks (m₁ = 1.0 kg, m₂ = 2.0 kg, and m3 = 3.8 kg) portrayed in the figure below move on a frictionless surface and a force F = 37 N acts as shown on the 3.8-kg block. m\ m₂ 1113 F Ⓒ (a) Determine the acceleration given this system. (b) Determine the tension in the cord connecting the 3.8-kg block and the 1.0-kg blocks. (c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block.

College Physics
11th Edition
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Chapter1: Units, Trigonometry. And Vectors
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Tutorial Exercise
Assume the three blocks (m₁ = 1.0 kg, m₂ = 2.0 kg, and m3 = 3.8 kg) portrayed in the figure below move on a frictionless surface and a force F = 37 N acts as shown on
the 3.8-kg block.
MI
a
=
m₂
(a) Determine the acceleration given this system.
(b) Determine the tension in the cord connecting the 3.8-kg block and the 1.0-kg blocks.
(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block.
ΣΕ.
m total
1113
Stop 1
(a) If we consider the system to consist of all three blocks plus the connecting rope, the total mass of the system is
Mtotal = M₁ + m₂ + m3 = 16.8
kg.
The only horizontal external force acting on this system is the applied force directed toward the right, which is chosen as the positive x-direction. Thus, Newton's second law
gives
kg
F
1 kg - m/s²
1 N
℗
=
m/s².
Transcribed Image Text:Tutorial Exercise Assume the three blocks (m₁ = 1.0 kg, m₂ = 2.0 kg, and m3 = 3.8 kg) portrayed in the figure below move on a frictionless surface and a force F = 37 N acts as shown on the 3.8-kg block. MI a = m₂ (a) Determine the acceleration given this system. (b) Determine the tension in the cord connecting the 3.8-kg block and the 1.0-kg blocks. (c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block. ΣΕ. m total 1113 Stop 1 (a) If we consider the system to consist of all three blocks plus the connecting rope, the total mass of the system is Mtotal = M₁ + m₂ + m3 = 16.8 kg. The only horizontal external force acting on this system is the applied force directed toward the right, which is chosen as the positive x-direction. Thus, Newton's second law gives kg F 1 kg - m/s² 1 N ℗ = m/s².
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