A block 0.1 m square, with 5 kg mass, slides down a smooth incline, 300 below the horizontal, on a film of SAE 30 oil at 200C that is 0.20 mm thick. If the block is released from rest at t = 0, what is its initial acceleration? Derive an expression for the speed of the block as a function of time. Plot the curve for V(t). Find the speed after 0.1 s. If we want the block to instead reach a speed of 0.3 m/s at this time, find the viscosity µ of the oil we would have to use.

a. acceleration (a) = 4.9 m/s²

b.  

c. see the attached diagram below

d  V = 0.048 m/s

e. The viscosity u = 0.27 Ns/m²

Explanation:

Given that :

m = 5 kg

A = ( 0.1 m )²

d = 0.20 mm

θ = 30 °

By applying Newton's second Law ; we have

F = mgsinθ -       (since no friction ; then

ma = mg sinθ

a = gsin θ

a = 9.8 × sin 30

a = 4.9 m/s²

b) By applying Newton's second law at any instant .

ma = mgsin θ -

and = T×A = -

Also; ma =

∴ ma = mg sin θ -

separating the variables ; we have

By integrating using limits ; we have:

∴

c) The curve for V(t) plot is attached in the diagram below.

d)

To find the speed after 1 sec; we have:

at t = 1.0 sec;

V =

V = 0.048 m/s

e) In order to determine the viscosity for which V(0.1) = 0.3 m/s ; we have the following.

∴

The viscosity u = 0.27 Ns/m²

you get 2.3=1.21*\frac{a}2 or 4.6=1.21a or a=3.802 (b)

explanation:

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