A 28.0 kg block is connected to an empty 2.00 kg bucket by a cord running over a

Question

A 28.0 kg block is connected to an empty 2.00 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.48 and the coefficient of kinetic friction between the table and the block is 0.33. Sand is gradually added to the bucket until the system just begins to move (Figure 1) .

Figure 1 link = https://session.masteringphysics.com/problemAsset/1000307676/1/GIANCOLI.ch05.p088.jpg

Part A

Calculate the mass of sand added to the bucket.

I got the correct answer which is 11. However, I can't get part B.

Part B

Calculate the acceleration of the system.

Express your answer using two significant figures.

a = __ m/s^2

Explanation / Answer

Static Friction , us = 0.48
Kinetic Friction , uk = 0.33
mB = 2.0 Kg
Let Mass of Sand = mS

Part A -
Minimum Force needed to just move the system should be equal to overcome Frcition Force.
Therefore,
Fr = us * N
Fr = 0.48 * m*g
Fr = 0.48 * 28 * 9.8 N
Fr = 131.72

This Force = Weight of Bucket
Wb = 131.72 N
(mB + mS ) * 9.8 = 131.72 N
mS = 131.72/ 9.8 - 2
mS = 11.44 Kg.

Part (B)
Now, Force Applied remains the same, but Friction Chamges From Static to Kinetic Friction.

Therefore, Frcition Force Reduces, which causes acceleration.

New Friction Force ==>
Fr = uk * N
Fr = 0.33 * m*g
Fr = 0.33 * 28 * 9.8 N
Fr = 90.55 N

Let the acceleration of the System = a
Now,
(11.44 + 2) * a = (11.44) * g - T -----------1
Where -
T - Fr = 28*a
T = 28 *a + 90.55 N

Substituting Value of T in eq 1
(11.44 + 2) * a = (11.44) * g - (28 *a + 90.55)
(11.44 + 2) * a = (11.44) * 9.8 - (28 *a + 90.55)
Solving for a

a = 0.52 m/s^2

Acceleration of the system, a = 0.52 m/s^2

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