A 4.0- kg wooden block rests on a level table. The coefficient of friction betwe

Question

A 4.0- kg wooden block rests on a level table. The coefficient of friction between the block and the table is 0.18. A 5.0- kg mass is attached to the block by a horizontal string passed over a frictionless pulley of negligible mass. Now, the 5.0- kg mass is released and the whole system accelerates.

What is the acceleration of the wooden block?

What is the tension in the string during the acceleration?

Determine the acceleration for the above situation when the coefficient of friction between the block and the table is 0.32.

What is the tension when the coefficient of friction between the block and the table is 0.32?

Explanation / Answer

Part A)

For the block hanging over the table

mg - T = ma

5(9.8) - T = (5)(a)

49-T = 5a

Solveing for T, we have

T = 49 - 5a

For the block on the table

T - Ff = ma

T - (umg) = ma

T - (.18)(4)(9.8) = 4a

T - 7.056 = 4a

Sub in for T

49 - 5a - 7.056 = 4a

a = 4.66 m/s^2

Part B
T = 49 -(5)(4.66)

T = 25.7 N

Part C)

For the block on the table now

T - Ff = ma

T - (umg) = ma

T - (.32)(4)(9.8) = 4a

T - 12.544 = 4a

Sub in for T

49 - 5a - 12.544 = 4a

a = 4.05 m/s^2

Part D)

T = 49 - (5)(4.05)

T = 28.8 N

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