A block with mass m_1 =7.1 kg rests on the surface of a horizontal table which h

Question

A block with mass m_1 =7.1 kg rests on the surface of a horizontal table which has a coefficient of kinetic friction of mu_k = 0.63. A second block with a mass m_2 = 10.9 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, sigma F_y, for block 2. Using the variable T to represent tension, write an expression for the sum of the forces in the x direction, sigma F_x for block 1. Write an expression for the magnitude of the acceleration of block 2, a_2, in terms of the acceleration of block 1, a_1. (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

m2 *g - T = m2*a for block 2

T - u*m1*g = m1*a for block 1

now adding the abouve two equation

we get a = (m2*g-u*m1*g )/(m1+m2) now substitute the values

a = 3.499 m/s^2

now keep the value of a in any one equation in above the we get T

lets take equation for block 2 and keep value a in that

m2*g - T = m2*(3.499 )

T = 68.67 N

(c) since string is ideal which means non stretcahble and not loose soo

the accelration magnitude of both bodies same

magnitude of accelration of block1 (a1) = mangitude of accelration of block2 (a2) = a = 3.499 m/s^2

but direction different

for block 2 in downward direction and block 1 right direction

(d) T = (m2*g - m1*a)/(m1+m2) [from block 2 equation]= m1*a + u*m1*g[block1 equation]

hope u got my complete explanaation

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