Question

a body slides down from rest from the top of 6.4m long rough plane inclined at 30degree with the horizontal find the time taken by block in reaching the bottom of the plane Take coefficient of friction to be 0.2 and g=9.8m/s^2

Answer 1

acceleration of block due to gravity on the plane is given by;

a = g SinΘ - f / m

a = g SinΘ - µR CosΘ / m

a = g SinΘ - µmg CosΘ / m

a = 9.8 x Sin 30 - µg CosΘ

a = 4.9 - (0.2)(9.8) Cos (30)

a = 4.9 - 1.697

a = 3.20 m/s^2

t = ?
If S = 6.4m

S = Ut + 1/2at^2 .................. U = initial velocity = 0 at the top of inclined plane

6.4 = 0 + 1/2(3.2)t^2

6.4 = 1.6t^2

=> t^2 = 6.4 / 1.6
=> t^2 = 4
=> √ t^2 = t = 2 seconds