A block with mass m_1 = 8.8kg rests on the surface of a horizontal table which h

Question

A block with mass m_1 = 8.8kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of mu k = 0.56. A second block with a mass m_2 = 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-directioa. sum F_y for block 2. Part Using the variable T to represent tension, write an expression for the sum of the forces in the x direction. If4 for block 1. part Write an expression for the magnitude of the acceleration of block 2. a3, in terms of the acceleration of block

Explanation / Answer

here,

m1 = 8.8 kg

uk = 0.56

m2 = 9.8 kg

a)

let the tension be T

for block 2,

sigma Fy = m2 * g - T

b)

for block 1,

sigma Fx = T - uk*m1 * g

c)

if the cable connecting the blocks is ideal

then the accelration of both the blocks are same

a1 = a2 = a

d)

accelration , a = net force /( effective mass)

a = ( m2 * g - uk * m1 * g) / (m1+ m2)

a = ( 9.8 * 9.8 - 0.56 * 8.8 * 9.8) /( 8.8 +9.8)

a = 2.57 m/s^2

for m2,

m2 * g - T = m2 * a

9.8 * 9.8 - T = 9.8 * 2.57

T = 121.23 N

the tension in the string is 121.23 N

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