Two blocks are connected by a cord passing over a small (frictionless) pulley as

Question

Two blocks are connected by a cord passing over a small (frictionless) pulley as shown below. The angle Theta = 30 degrees, and the mass of the small block is m = 15 kg. If the coefficient of friction between the small block and the inclined plane is 0.25 and the large block is accelerating down at 2 m/s2 , what is the mass, M, of the large block?

A. 17.35 kg

B. 9.34 kg

C. 15 kg

D. 73.50 kg

E. 11.47 kg

Two blocks are connected by a cord passing over a small (frictionless) pulley as shown below. The angle Theta = 30 degrees, and the mass of the small block is m = 15 kg. If the coefficient of friction between the small block and the inclined plane is 0.25 and the large block is accelerating down at 2 m/s2 , what is the mass, M, of the large block? A. 17.35 kg B. 9.34 kg C. 15 kg D. 73.50 kg E. 11.47 kg

Explanation / Answer

Let T be the tension in string

Applying newton second law on M

Mg - T = M*2.................(1)

we need value of T

since string length is constant it means m is acceleratin 2m/s^2 up the inclined plane

Now Applying newton second law on m

T - f - mgsinx = ma

T - 0.25*15*10*cos30 - 15*10*sin30 = 15*2

we get T = 137.46 N

putting it in (1)

we get

M= 17.1844 Kg

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