I Help Homework 4 Begin Date: 2/9/2018 12:00:00 AM Due Date: 2/26/2018 11:59:00

Question

I Help Homework 4 Begin Date: 2/9/2018 12:00:00 AM Due Date: 2/26/2018 11:59:00 PM End Date: 3/26/2018 11:59:00 PM (7%) Problem 21: A block with mass m,-8.5 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction ofPk = 0.73. A second block with a mass m2 = 10.3 kg is connected to the first by an ideal pulley system such that the second block is hanging vertically. The second block is released and motion occurs. theexpertta.com 20% Part (a) Using the variable T to represent tension, write an expression for the sum of the forces in the y-direction, , for block 2 Grade Summary 100% Attempts remaining: (0% per attempt) detailed view 0 I give up! deduction per feedback. Hint Hints: 2% deduction per hint. Hints remaining Feedback: 2 - 20% Part (b) Using the variable T to represent tension, write an expression for the surn of the forces in the x direction,2F for block 20% Part (c) Write an expression for the magnitude of the acceleration of block 2, a2 in terms of the acceleration of block 1 , a, (Assume the cable connecting the masses is ideal.) 20% Part (d) Write an expression using the variables provided for the magnitude of the tension force, T 20% Part (e) What is the tension, T in Newtons?

Explanation / Answer

here,

a)

let the tension in the cable be T

for block 2 ,

equating the forces in y-direction

sigma(Fy) = m2 * g - T

b)

for horizontal direction

sigma(Fx) = T

c)

as the pulley is massless and frictionless

accelration , a = a1 = a2

d)

sigma(Fy) = m2 * g - T = m2 * a ...(1)

and

sigma(Fx) = T = m1 * a ...(2)

from (1) and (2)

T = m1 * m2 * g /( m1 + m2)

e)

m1 = 8.5 kg

m2 = 10.3 kg

uk = 0.73

T = 8.5 * 10.3 * 9.81 /( 8.5 + 10.3) N

T = 45.7 N

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