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(14%) an ideal (frictionless and massless) pulley such that one block with mass m 13.75 kg is on a horizontal table and the other block with mass m-9.5 kg hangs vertically. Both blocks experience gravity and the tension force, T. Use the coordinate system specified in the diagram. Problem 7: Two blocks are connected by a massless rope, The rope passes over Otheexpertta.com 25% Part (a) Assuming friction forces are negligible, write an expression using only the variables provided for the acceleration that the block of mass m/ experiences in the x-direction. Your answer should involve the tension, T Grade Summary Deductions 0% Potential 100% - Submissions 4 5 6 Attempts remaining: 즈 (2% per attempt) detailed view IT 0. Submit | Hint I give up! Hints: 1% deduction per hint. Hints remaining: 3 Feedback: 0 dedaction per foodback ▲ 25% Parti b) Under the same assumpuons, write an expression for the acceleration, a. the block of mass m expenences the v direction.
Otheexpertta.com 25% Part a) Assuming friction forces are negligible write an expression, using only the variables provided for the acceleration that he block of mass mi cxperiences in the x-direction. Your answer should involve the tension, T Grade Summary Deductions 0% al = Potential 100% 7 8 9 Submissions Attempts remaining:8 (2% per attempt) detailed view 1 2 3 m. 0 m2 Submit Hint I giveup!」 deduction per feedback. Hints: 1% dedaction per hint. Hints remaining:- Feedback: 呈▲ 25% Part (b) Under the same assumptions, write an expression for the acceleration, as the block of mass m2 experiences in they direction Your answer should be in terms of the tension, T and m2 ▲ 25% Part (c) Carefully consider how the accelerations ai and az are related. Solve for the magnitude of the acceleration, ai, of the block of mass mi, in meters per square second. 呈厶25% Part (d) Find the magnitude of the tension in the rope, T, in newtons
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Answer #1

a) considering equilibrium of forces along horizontal

T = m1 a1

a1 = T / m1

=======

b) considering equilibrium of forces along vertical

m2 a2 = m2 g - T

a2 = - ( m2 g - T) / m2

=======

c)

since the system attached to a same string so acceleration must be same

m2 a = m2 g - m1 a

a = m2 g / ( m1 + m2)

a = 9.5* 9.8 / ( 9.5 + 13.75)

a = 4 m/s^2

=======

d)

T= m1 a = 13.75 * 4

T = 55 N

========

Comment in case any doubt, will reply for sure.. Goodluck

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