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

Question was not very clear if there are specific answer choices tell me and i'll try to you more. hope this though2 laps is 1 min. we know that for each min she does two times the amount of laps. for 12 laps we can do 12/2 which does give us 12 laps in 6 mins not 16 mins.

Answer: buoyancy forceexplanation: in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.