Introductory Physics - Systems

Question

Hello, I have attempted this question multiple times but do not know how to answer it, please help me!

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

Hi Rahman:

The mass of the full bucket = 0.5 + 2.1 = 2.6 kg, so its weight (acting vertically down) = 2.6 x 9.8 Newton

The force of static friction between the block and the desk resists the full bucket's weight

The force of static friction on the block is F_s = μ_s x N [μ_s= coefficient of static friction, N = normal force]

The normal reaction on the block is equal to its weight = (6.5 x 9.8) Newton

Just before the block starts to move, these 2 forces are in equilibrium, and so bucket weight, W = F_s

Therefore, 6.5 x 9.8 μ_s = 2.6 x 9.8

Dividing both sides by 9.8 gives μ_s = 2.6 / 6.5 = 0.4

Answer, coefficient of static friction, μ_s, = 0.4