Ablock of mass m1 = 280 g is at rest on a plane that makes an angle θ = 30° above the horizontal. the coefficient of kinetic friction between the block and the plane is μk = 0.10. the block is attached to a second block of mass m2 = 200 g that hangs freely by a string that passes over a frictionless and massless pulley. find its speed when the second block has fallen 30.0 cm.

The speed is 1.016 m/s or 102 cm/s

Explanation:

Step 1: Data given

m1 = 0.280 kg

m2 = 0.200 kg

θ = 30°  

μk = 0.10

Δy = -30 cm = -0.30 m

Step 2

m has tension T pulling up, opposed by gravity (-mg) pulling down.  The acceleration is vertically down, so = -a

T - mg = -ma ⇒ T - 0.200(9.81) = -0.200a ⇒

T - 1.962 = -0.2a

T = -0.2a +1.962

m1 has the same tension T pulling up the plane, and opposed by -m1*g*sinθ and -μk*m1*g*cosθ (pointing down the plane).  

Its acceleration is +a (up the plane).  This can be given as:

T - m1*g*sinθ - μk*m1*g*cosθ = m1*a ⇒

T - 0.280*9.81*0.5 - 0.1*0.280*9.81*0.866 = 0.28a ⇒

T - 1.135527 = 0.28a

since T = -0.2a +1.962 we can eliminate T as followed

-0.2a +1.962 -1.135527 = 0.28a

0.826473 = 0.48a

a = 1.722 m/s²

Kinematic relation for m:

 v²- 0² = 2(-a)Δy = 2(-1.722)(-0.30) (m/s)2

 v = √1.0332 m/s ≅ 1.016 m/s = 102 cm/s

The speed is 1.016 m/s or 102 cm/s