Question

A 5kg block begins from rest and slides down on an inclined plane. After 4s, the block has a velocity of 6m/s. If the angle of inclination of the plane is 45 degrees, approximately how far has the block traveled after 4s

Answer 1

Since the block is not going vertically so it will not experience acceleration = g

its acceleration would be gsinθ

now given

m = 5 kg

u = 0

t = 4 sec

v = 6m/s

θ = 45⁰

a = $10*\frac{1}{\sqrt{2} }$

s = ?

using the equation

s = ut + $\frac{1}{2} at^{2}$

s = 0(4) + $\frac{1}{2} (10)(\frac{1}{\sqrt{2} } ) 4^{2}$

 =$5*\frac{1}{\sqrt{2} } *16$

= 40$\sqrt{2}$ m

Please note that here in calculation we did not use mass of the body. In fact any mass a small stone or Himalaya will go for same distance until and unless we consider the existence of friction and wind resistance.

Hope it helps :)