O fs mg cos 30° O 02 points I Provo us Answers Ask Your In the figure below, m1

Question

O fs mg cos 30° O 02 points I Provo us Answers Ask Your In the figure below, m1 3.7 kg, m2 5.1 kg, and the coefficient of kinetic friction between the inclined plane and the 3.7-kg block is An- 0-3, Find the magnitude of the acceleration of the masses and the tension in the cord. x m/s? X N Choose a coordinate system in which the x direction is up the incline for the block on the incline and downward for the hanging block. Draw free-body dlagrams or the four forces acting on the block on the incline and the two forces acting on the hanging block. Because they are attached by the string, the blocks will have a common acceleration. Apply Newton's second law to each block separately Solve the resulting expressions simultaneously for both the acceleration and tension. Isutoma Arewer save Progress Practice Another Version Mr Notes o Ask Your Teacher 5. o points Previous Answers Tholens 5 PO57.

Explanation / Answer

here

Fnet = m * a

Fnet = 5.1 *a = W - T

W - T = m * g - T = 5.1 * 9.8 - T

5.1 * a = 49.98N - T

T = 49.98 - 5.1 * a ..........................................................(1)

Now, for m1
gravity parallel to plane F * g = m * g * sin(theta) = 3.7 * 9.8 * sin30 = 18.13 N

Ff = u * m * g * cos(theta) = 0.3 * 3.7 * 9.8m/s² * cos30 = 9.42 N

then tension is

Fnet = m * a = 3.7 * a = T - Ff - Fg = T - 18.13N - 9.42N = T - 27.55N

T = 3.7 * a + 27.55 .......................................(2)

from 1 and 2 equation we get

Since T = T,

49.98 - 5.1 * a= 3.7 * a + 27.55

a = 2.54 m/s^2

so the acceleration is 2.54 m/s^2

then the tension is

T = 49.98 - 5.1 * 2.54 = 37.026 N

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