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?

Explanation

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.