Question

A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3⋅2 m high. How much work was required (a) to lift the block from the ground and put it an 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 to the ground (d) it slides down the incline? Take g = 10 m/s².

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{block},m=200g=0.2\mathrm{kg} \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{incline},s=10\mathrm{m}, \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{incline},h=3.2\mathrm{m} \\ \mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity},g=10\mathrm{m}/s^2 \end{gathered}$$

(a)
Work done, W = mgh = 0.2 × 10 × 3.2 =6 .4 J

(b)
Work done to slide the block up the incline
$$\begin{gathered} \mathrm{W}=\left(mg\sin \theta \right)\times \mathrm{s} \\ =\left(0.2\right)\times 10\times \left(3.2/10\right)\times 10 \\ =6.4\mathrm{J} \end{gathered}$$

(c)
Let final velocity be v when the block falls to the ground vertically.
Change in the kinetic energy = Work done
$\frac{1}{2}m{v}^2-0=6.4\mathrm{J}$
$\Rightarrow \nu =8\mathrm{m}/s$

(d)
Let $\nu$ be the final velocity of the block when it reaches the ground by sliding.
$\frac{1}{2}m{\nu }^2-0=6.4\mathrm{J}$
$\Rightarrow v=8\mathrm{m}/s$