A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 32 in high. How much work was required (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground, if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline? Take g = 10 m/s2.
Mass of the block, m = 200 g = 0.2 kg
Acceleration due to gravity, g = 10 m/s2
Length of the incline plane, L = 10 m
Height of the incline plane, h = 3.2 m
(a.) Required work for lifting the block to the top of the incline = mgh
= 0.2×10×3.2
= 6.4 J
(b.) Work done against friction = 0 J (as the surface is frictionless)
Required work for sliding the block up = Work done against gravity
= 6.4 J
(c.) Let the speed of the block when it falls off the incline be v m/s.
The speed of the block when it is at rest on the top, u = 0 m/s
According to equation of motion, v2 - u2 = 2gh
⇒ v2 - 0 = 2×10×3.2
⇒ v2 = 64
⇒ v = 8 m/s
(d.) When the block slides down, let the speed of the block be v m/s
Change in kinetic energy of the block = work done by gravity + work done by friction
⇒ 1/2 mv2 = mgh + 0
⇒ v2 = 2gh
⇒ v2 = 2×10×3.2 = 64
⇒ v = 8 m/s