A block with mass mi = 9.1 kg rests on the surface of a horizontal table which h

Question

A block with mass mi = 9.1 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction oiuk = 0.69. A second block with a mass m2 = 9.5 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. Using the variable T to represent tension, write an expression for the sum of the forces in the y-direction, ZF_ys for block 2. Part (b) Using the variable T to represent tension, write an expression for the sum of the forces in the x direction, ZF_X for block 1. Write an expression for the magnitude of the acceleration of block 2, 02, in terms of the acceleration of block 1 ,aj. (Assume the cable connecting the masses is ideal.) Write an expression using the variables provided for the magnitude of the tension force, T. What is the tension, T in Newtons?

Explanation / Answer

a) net force on block2, Fnety = m2*g - T

b) net force on block1, Fnetx = T

c) As the blocks are connected with a string they both have same acceleration.

Fnety = m2*g - T

m2*a = m2*g - T

a = g - T/m2 ---(1)

Fnetx = T

m1*a = T

a = T/m1   --(2)

d) from equations 1 and 2

T/m1 = g - T/m2

T*(1/m1 + 1/m2) = g

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

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

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

= 9.1*9.5*9.8/(9.1+9.5)

= 45.5 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.