Physics Systems

Question

Hello, I have attempted this question multiple times but do not know how to answer it, please help me! Can anyone attempt part B please?

A 6.5kg block is at rest on top of a desk. it is connected with a rope over a pulley an empty water bucket with an (empty) mass 0.500 kg.

a)Water is added slowly to the bucket the split second before the system starts to accelerate it is noted that 2.1L (also kg) has been added to the bucket, Calculate the coefficient of static friction between 6.5kg block at the desk.

b)one extra drop (negligible mass)is added to the value calculated from part a, setting the system in motion. The system accelerates at a rate of 1.5m/s^2. what is the coefficient of kinetic friction between the 6.5 kg block and the table.

Answer 1

I'm guessing it looks like this:

a) T = W₂ = mg = [0.5 + (2.1L)(1kg/L)](9.81 m/s²) = 25.506 N

N = W₁ = mg = (6.5 kg)(9.81 m/s²) = 63.765 N

T = F_F = μN so μ = F_F/N = T/N = W₂/W₁ = 25.506/63.765 = 0.40

b) Look at the lower free body diagram only

Net Force = ma [where m is the mass of the lower bucket (2.6 kg) and a = -1.5 m/s²]

Net Force = T - W₂

W₂ = 25.506 N (from above)

T = μN so:

μN - W₂ = ma

μ = (ma + W₂)/N = [(2.6)(-1.5) + 25.506]/63.765 = 0.34