Block a, with a mass of 10 kg, rests on a 30° incline. The coefficient of kinetic friction is 0. 20. The attached string is parallel to the incline and passes over a massless, frictionless pulley at the top. Block b, with a mass of 3. 0 kg, is attached to the dangling end of the string. The acceleration of b is:.

Consult the attached free body diagram.

Use Newton's second law to set up several equations:

• net force on block B

∑ F = F[tension] - (3.0 kg) g = (3.0 kg) a

Note that we take the positive direction to be "upward".

Equivalently,

F[tension] = (3.0 kg) a + 29.4 N

• net force on block A parallel to the incline

∑ F = (10 kg) g sin(30°) - F[tension] ± F[friction] = (10 kg) a

Note that we take "down the incline" to be the positive direction.

The sign of the frictional force depends on which direction block A moves along the incline. To determine which way the block will slide, suppose for a moment that there is no friction so that the equation reduces to

49 N - F[tension] = (10 kg) a

If we combine this equation with the one we got from block B, we eliminate F[tension] and find

(F[tension] - 29.4 N) + (49 N - F[tension]) = (3.0 kg) a + (10 kg) a

⇒   (13.0 kg) a = 19.6 N

⇒   a ≈ 1.5 m/s²

Since this acceleration is positive, we know block A wants to slide down the incline. Then F[friction] points in the negative direction (up the plane), so that

∑ F = (10 kg) g sin(30°) - F[tension] - F[friction] = (10 kg) a

or more simply,

49 N - F[tension] - F[friction] = (10 kg) a

and upon substituting F[tension] from above,

49 N - ((3.0 kg) a + 29.4 N) - F[friction] = (10 kg) a

49 N - (3.0 kg) a - 29.4 N - F[friction] = (10 kg) a

F[friction] = 19.6 N - (13 kg) a

• net force on A perpendicular to the incline

∑ F = F[normal] - (10 kg) g cos(30°) = 0

It follows that

F[normal] = (10 kg) g cos(30°) ≈ 85 N

so that

F[friction] = µ F[normal] ≈ 0.20 (85 N) = 17 N

Substitute F[friction] and solve for the acceleration a :

17 N ≈ 19.6 N - (13 kg) a

(13 kg) a ≈ 2.6 N

a ≈ 0.20 m/s²

answer: yay points!

explanation: