A small block is attached on the left of a larger block by a thin layer of grease, 2 micrometres (10^-6) thick. The mass of the smaller block is 0.15 kg and the area of contact is 8x10^-4 metres squared. If the smaller block slides down 5 mm in 2 weeks, calculate the viscosity of the grease. ANSWER: 9.07x10^5 Pa.s

I basically need help in getting to the answer. A formula which may help is Force/Area = Viscosity(change in velocity)/distances between areas of contact F/A = (n x delta v)/delta x

Using your formula I get 8.89x 10^5 Pa-s.

Fiirst, the value of delta v is the average velocity of this sliding block, which is 5 mm/2 weeks, or 4.1336x10-9 m/s. I then rearranged your formula to this:

n = (F x delta x)/(A x delta v)

Note that F is the sliding force due to the weight of the block: F = 0.15 Kg * 9.8 m/s^2 = 1.47 N. So you have:

n = (1.47 N x 2x10^-6 m)/ (8x10^-4 m^2 x 4.1336x10-9 m/s) = 8.89x10^5 N-s/m^2 = 8.89x10^6 Pa-s.

The difference between my answer and yours is that I used g = 9.8 m/s^2, whereas your answer appears to have used g = 10 m/s^2. It drives me nuts when people round 9.8 to 10, and then show their answer to three significant digits!

Thank you so much! :D