Problem 7: In the figure below we have three blocks of mass m₁ = 2 kg, m₂ = 4 kg, and m3 = 10 kg. m₂ and m₂ are on an inclined plane with 8 = 30° and coefficient of kinetic friction 0.2. m₁ and m₂ are connected by a rope that provides a tension T₁. m₂ and m3 are connected by rope that provides a tension T₂ and this rope goes over a frictionless pulley. (a) What is the acceleration of the system? Answer: 3.7 m/sec². (b) What is the tension T₁ on the rope connection m, and m₂? Answer: 20.6 N. (c) What is the tension T₂ on the rope connection m₂ and m3? Answer: 61.0 N. T2 0 m1 T₁ Hk m₂ Hk T2 M3

University Physics Volume 1
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Chapter1: Units And Measurement
Section: Chapter Questions
Problem 81AP: Consider the equation s=s0+v0t+a0t2/2+j0t3/6+s0t4/24+ct5/120 , were s is a length and t is a time....
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Problem 7:
In the figure below we have three blocks of mass m₁ = 2 kg, m₂ = 4 kg, and m² = 10 kg. m₁
and m₂ are on an inclined plane with 0 = 30° and coefficient of kinetic friction 0.2. m₁ and m₂
are connected by a rope that provides a tension T₁. m₂ and m3 are connected by rope that
provides a tension T₂ and this rope goes over a frictionless pulley.
(a) What is the acceleration of the system? Answer: 3.7 m/sec².
(b) What is the tension T₁ on the rope connection m₁ and m₂? Answer: 20.6 N.
(c) What is the tension T₂ on the rope connection m₂ and m3? Answer: 61.0 N.
T₂
0
m1
T₁
Hk
M₂
Uk
T2
M3
Transcribed Image Text:Problem 7: In the figure below we have three blocks of mass m₁ = 2 kg, m₂ = 4 kg, and m² = 10 kg. m₁ and m₂ are on an inclined plane with 0 = 30° and coefficient of kinetic friction 0.2. m₁ and m₂ are connected by a rope that provides a tension T₁. m₂ and m3 are connected by rope that provides a tension T₂ and this rope goes over a frictionless pulley. (a) What is the acceleration of the system? Answer: 3.7 m/sec². (b) What is the tension T₁ on the rope connection m₁ and m₂? Answer: 20.6 N. (c) What is the tension T₂ on the rope connection m₂ and m3? Answer: 61.0 N. T₂ 0 m1 T₁ Hk M₂ Uk T2 M3
Fnet = ma, x = xo + Voxt + axt², vx = vox + axt, (vx)² = (vox)² + 2āxx, g = 9.8 m/sec²
Transcribed Image Text:Fnet = ma, x = xo + Voxt + axt², vx = vox + axt, (vx)² = (vox)² + 2āxx, g = 9.8 m/sec²
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