Part A A 5.2 kg box is on a frictionless 38 ° slope and is connected via a massl

Question

Part A A 5.2 kg box is on a frictionless 38 ° slope and is connected via a massless string over a massless frictionless pulley to a hanging 1.9 kg weight. What is the tension in the string if the 5.2 kg box is held in place, so that it cannot move? Express your answer using two significant figures Submit My Answers Give Up Part B If the box is then released, which way will it move on the slope? Upward Downward Submit My Answers Give Up Part C What is the tension in the string once the box begins to move? Express your answer using two significant figures Submit My Answers Give Up

Explanation / Answer

Here

T = mg

= 1.9*9.8

= 18.62 N

Here

Downward Force = MgSin(theta)

= 5.2*9.8*Sin(38 degree)

= 31.374 N

AS 31.374 N > 18.62 N

SoIt will Slide Downward along the inclined

Here

Acceleration= Net Force/TYotal Mass

= (31.374-18.62)/(5.2+1.9)

= 1.796m/sec^2

T - mg = ma

Therefore

T = 1.9*(9.8+1.796)

= 22.033 N

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