Physics question!?

Question

"Block A, with a mass of 35 kg, rests on a 65o incline. The coefficient of kinetic friction is μk = 0.3. An attached rope is parallel to the incline and passes over a massless, frictionless pulley at the top of the plane. Block B, with a mass of 9 kg, is attached to the dangling end of the rope. What is the acceleration of block B? Define down as the -y direction."
My solution was -2.0 m/s^2 but I got wrong. please help me!

Answer 1

I got the answer as - 5.06m/s^2. If it is correct, you may mail me at [e-mail] for the procedure.

Answer 2

Forces on Block B:

T (tension in rope, upward)
Weight (downward) = (9kg)•g

So, by Newton's 2nd law:

a = Fnet/(9kg) = ((9kg)•g–T)/(9kg)

We don't know what "T" is yet, but that's where the other block comes in.

Forces on Block A:

T (diagonally upward--same value as other T)
Weight (downward) = (35kg)•g
Fn (normal force) = Weight • cos(65) = (35kg)•g•cos(65)
Friction (diagonally downward) = Fn•Î¼k = (35kg)•g•cos(65)•Î¼k

Break the Weight vector into 2 components
* Parallel to slope: (35kg)•g•sin(65)
* Perpendicular to slope: (35kg)•g•cos(65)

The perpendicular component cancels out with the Normal force. So the Net Force on Block B is:

T–(35kg)•g•sin(65) – (35kg)•g•sin(65)•Î¼k
= T–(35kg)•g•sin(65)•(1+μk)

and its acceleration is:

a = (T–(35kg)•g•sin(65)•(1+μk)) / 35kg

Now note that this "a" is the same value as the other "a" (since the blocks are connected, they must accelerate at the same rate). So solve both equations simultaneously; that will give you the values of T and a.

Answer 3

♠ 2nd Newton says: net force F=m*a, where F=-f1+f2-f3, m=m1+m2, a is acceleration to be found, m1=9 kg, m2=35 kg, f1=m1*g pulling back, f2=m2*g*sin65° the component of weight of block A along the incline down, f3=μ*m2*g*cos65° is friction force due to component of weight of block A, normal to the incline; Thus
♣ -m1*g +m2*g*(sin65 - μ*cos65) = (m1+m2)*a, hence
a=g*(-m1 + m2*(sin65 - μ*cos65)) / (m1+m2) =4.072 m/s^2; block B going up!