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

(a) The work required to lift the block from the ground and put it an the top is $6.40 \mathrm{J}$

(b) The work required to slide the block up the incline is $6.40 \mathrm{J}$

(c) When the block falls vertically on the ground, the velocity will be $8 m / 8$

(d) It slides down the slope at a velocity of 8 m / s

Explanation:

Given values  :-

Block mass = 200 g = 0.2 kg

Length of the  Incline, I = 10 m.

Height of the frictionless incline, h = 3.2 m

(a) Now the work required is equal to the work done to combat the gravity.

W = mgh

$W=0.2 \times 10 \times 3.2$

$\mathrm{W}=6.40 \mathrm{J}$

(b) Since here the slope is frictionless, gravity is the only force that works here.  

Therefore, the work done against gravity is the same in this too.

W= mgh  = 6.40 J

(c) When the block falls vertically on the ground:- speed

$\begin{aligned}&v^{2}=u^{2}+2 g h\\&v^{2}=0+2 \times 10 \times 3.2\\&v^{2}=64\\&\mathrm{v}=8 \mathrm{m} / \mathrm{s}\end{aligned}$

When it falls down the ground,  Only force acting on the block is gravity, so the change in Kinetic energy is equal to gravity function.  

(d)  Work done by gravity  = Change in Kinetic energy  

$\mathrm{mgh}=\frac{1}{2}\left(\mathrm{mv}^{2}\right)$

$v^{2}=2 g h$

$v^{2}=\sqrt{2 \times 10 \times 3.2}$

$v^{2}=\sqrt{20 \times 3.2}$

$v=\sqrt{64}$

$\mathrm{v}=8 \mathrm{m} / \mathrm{s}$

It slides down the slope at a velocity of 8 m / s