- #1

Benjamin_harsh

- 211

- 5

- Homework Statement
- A block of mass M is released from point P on a rough inclined

plane with inclination angle θ, shown in the figure below. The coeffucient

of friction is μ. if μ < tan θ, then time taken by the block to reach another

point Q on the inclined plane, where PQ = s, is

- Relevant Equations
- What is the meaning of " PQ = s" in the question?

How ##t = \large\sqrt\frac{2s}{a} ##?

Sol:

##Mg.sinθ - μMg.cosθ = ma##

##a = g.sinθ - μg.cosθ##

Now ##S = ut + \large\frac{1}{2}\normalsize at^{2}##

but ##u = 0##

##t = \large\sqrt\frac{2s}{a} = \sqrt\frac {2s}{gcosθ(tanθ- μ)}##

What is the meaning of " PQ = s" in the question?

How ##t = \large\sqrt\frac{2s}{a} ##?

How ##a =gcosθ(tanθ- μ) ## in that last square root step?

##Mg.sinθ - μMg.cosθ = ma##

##a = g.sinθ - μg.cosθ##

Now ##S = ut + \large\frac{1}{2}\normalsize at^{2}##

but ##u = 0##

##t = \large\sqrt\frac{2s}{a} = \sqrt\frac {2s}{gcosθ(tanθ- μ)}##

**My questions:**What is the meaning of " PQ = s" in the question?

How ##t = \large\sqrt\frac{2s}{a} ##?

How ##a =gcosθ(tanθ- μ) ## in that last square root step?