4. Two blocks connected by a rope of negligible mass are being pulled by a horizontal force F. m2 a. Draw free-body diagrams for the two masses. Given that m, = 14.0 kg, m2 = 26.0 kg, the coefficient of kinetic friction between each block and the horizontal surface is µg = 0.25, and that the magnitude of the force F is 200 N, b. Find the tension force, T, in the rope.
4. Two blocks connected by a rope of negligible mass are being pulled by a horizontal force F. m2 a. Draw free-body diagrams for the two masses. Given that m, = 14.0 kg, m2 = 26.0 kg, the coefficient of kinetic friction between each block and the horizontal surface is µg = 0.25, and that the magnitude of the force F is 200 N, b. Find the tension force, T, in the rope.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
#4 using any of these equations
![4. Two blocks connected by a rope of negligible mass
are being pulled by a horizontal force F.
m1
m2
а.
Draw free-body diagrams for the two
masses.
Given that m, = 14.0 kg, m2 = 26.0 kg, the coefficient of kinetic friction between each
block and the horizontal surface is µg = 0.25, and that the magnitude of the force F is
200 N,
b. Find the tension force, T, in the rope.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc69edb92-08ea-4b49-88be-8d812bc0041d%2F60a95929-6a38-4ab9-af1f-3b567f49d2ae%2Fslfd2r8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4. Two blocks connected by a rope of negligible mass
are being pulled by a horizontal force F.
m1
m2
а.
Draw free-body diagrams for the two
masses.
Given that m, = 14.0 kg, m2 = 26.0 kg, the coefficient of kinetic friction between each
block and the horizontal surface is µg = 0.25, and that the magnitude of the force F is
200 N,
b. Find the tension force, T, in the rope.
![Kinematics:
Vx = Vxo + axt
(x – xo) = vxot +;axt?
vž – vžo = 2a,(x – xo)
Vy = Vyo + Ayt
(y – yo) = vyot +ayt?
vž – vžo = 2a,(y – yo)
1
%3D
7 = xî + yĵ
di
う=
Vzî + vyĵ
dt
axî + ayĵ
dt
E =
= m
dt
di
(constant mass)
dt
EF = må, (constant mass)
Kinetic Friction Force:
fk = HxN
Static Friction Force: fs < HsN
g = 9.80 m/s²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc69edb92-08ea-4b49-88be-8d812bc0041d%2F60a95929-6a38-4ab9-af1f-3b567f49d2ae%2F0sj4ehl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Kinematics:
Vx = Vxo + axt
(x – xo) = vxot +;axt?
vž – vžo = 2a,(x – xo)
Vy = Vyo + Ayt
(y – yo) = vyot +ayt?
vž – vžo = 2a,(y – yo)
1
%3D
7 = xî + yĵ
di
う=
Vzî + vyĵ
dt
axî + ayĵ
dt
E =
= m
dt
di
(constant mass)
dt
EF = må, (constant mass)
Kinetic Friction Force:
fk = HxN
Static Friction Force: fs < HsN
g = 9.80 m/s²
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