The velocity of a 2.5 kg block sliding down an inclined plane ( mu = 0.2) is found to be 1.5 "m.s"^(-1). One second later it has a velocity of 5 "m.s"^(-1). What is the angle of the plane with respect to the horizontal?

The velocity of a 2.5 kg block sliding down an inclined plane ( mu = 0

1.	The velocity of a 2.5 kg block sliding down an inclined plane ( mu = 0.2) is found to be 1.5 "m.s"^(-1). One second later it has a velocity of 5 "m.s"^(-1). What is the angle of the plane with respect to the horizontal?
Answer» SOLUTION :Downward resultant force on the BLOCK along the inclined plane = mg sin `theta = muN = mgsintheta = mu COS theta` = mg (sin `theta -mu cos theta`) Downward acceleration, a = g(sin`theta -MUCOS theta) =9.8 (sin theta - 0.2 cos theta ) "m.s"^(-1)` From the relation v = u+at, a = `(v-u)/(t) = (1-1.5)/(1) = 3.5 "ms."^(-2)` `:. 9.8 (sin theta -0.2 cos theta) ` = 3.5 or, sin `theta-0.2 cos theta = (3.5)/(9.8) = (5)/(14)` Solving this equation we get `theta = 32^(@)` It is the angle of the plane with respect to the horizontal.