A block of mass m1 = 21.2 kg is at rest on a plane inclined at = 30.0° above the

Question

A block of mass m1 = 21.2 kg is at rest on a plane inclined at = 30.0° above the horizontal. The block is connected via a rope and massless pulley system to another block of mass m2 = 28.0 kg, as shown in the figure. The coefficients of static and kinetic friction between block 1 and the inclined plane are s = 0.109 and k = 0.086, respectively. If the blocks are released from rest, what is the displacement of block 2 in the vertical direction after 1.89 s? Use positive numbers for the upward direction and negative numbers for the downward direction.

Explanation / Answer

Here,

let the acceleration of the blocks is a

a = net force/total mass

a = (m2 * g - m1 * g * sin(theta) - u * m1 *g *cos(theta) )/(m1 + m2)

a = (28 - 21.2 * sin(30) - 21.2 * 0.109 * cos(30)) * 9.8/(28 + 21.2)

a = 3.07 m/s^2

the acceleration is 3.07 m/s^2

for the block 2 ,

velocity , v = - 3.07 * 1.89

v = - 5.802 m/s

the velocity of block 2 is - 5.802 m/s

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