Problem 4.33 A block with mass m A = 14.0 kg on a smooth horizontal surface is c

Question

Problem 4.33

A block with mass mA = 14.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second block with mass mB = 4.0 kg which hangs vertically. (Figure 1)

Part A

Determine the magnitude of the acceleration of the system.

Express your answer to two significant figures and include the appropriate units.

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Part B

If initially mA is at rest 1.250 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely?

Express your answer to two significant figures and include the appropriate units.

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Part C

If mB = 1.0 kg , how large must mA be if the acceleration of the system is to be kept at g100 ?

Express your answer to two significant figures and include the appropriate units.

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Figure 1 of 1

Problem 4.33

A block with mass mA = 14.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second block with mass mB = 4.0 kg which hangs vertically. (Figure 1)

Part A

Determine the magnitude of the acceleration of the system.

Express your answer to two significant figures and include the appropriate units.

SubmitHintsMy AnswersGive UpReview Part

Part B

If initially mA is at rest 1.250 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely?

Express your answer to two significant figures and include the appropriate units.

a =

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Part C

If mB = 1.0 kg , how large must mA be if the acceleration of the system is to be kept at g100 ?

Express your answer to two significant figures and include the appropriate units.

t =

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mA =

Figure 1 of 1

Explanation / Answer

A>

motive force Fm = mb*g = 4*9.806 = 39.224 N
acceleration a = Fm/(ma+mb) = 39.224/18 = 2.17 m/sec^2

B.

By s = ut + 1/2at^2
=>1.25 = 0 + 1/2 x 2.17 x t^2
=>t = 1.15
=>t = 1.07 sec
C.

By T = mA x 0.098 --------------(i)
& 1g - T = 1 x 0.098 -----------(ii)
By (i) + (ii) :-
=>9.8 = 0.098mA + 0.098
=>mA = 99 kg

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