Question

A ball is released from the top of a tower. The ratio of work done by force of gravity in first, second and third second of the motion of the ball is [a] 1 : 2 : 3 [b] 1 : 4 : 9 [c] 1 : 3 : 5 [d] 1 : 5 : 3

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Ball is released Therefore Initial velocity (u) = 0 m/s.
• Acceleration = g.

$\huge{\bold{\underline{Explanation:-}}}$

Let the Distances travelled in 1st, 2nd and 3rd seconds be h₁ ,h₂ and h₃.

Now, From Work done Formula,

$\large{\boxed{\tt W = mgh}}$

The work done in 1st second,

$\large{\tt \leadsto W_1 = mgh_1}$

$\large{\tt \leadsto W_1 = mgh_1 \: ----(1)}$

The work done in 2nd second,

$\large{\tt \leadsto W_2 = mgh_2}$

$\large{\tt \leadsto W_2 = mgh_2 \: ----(2)}$

The work done in 3rd second,

$\large{\tt \leadsto W_3 = mgh_3}$

$\large{\tt \leadsto W_3 = mgh_3 \: ----(3)}$

Now, Taking The ratio's

I.e $\large{\bold{\tt W_1 : W_2 : W_3 }}$

Substituting the values,

$\large{\tt \leadsto W_1 : W_2 : W_3 = mgh_1 : mgh_2 : mgh_3}$

$\large{\tt \leadsto W_1 : W_2 : W_3 = (mg) \times h_1 : (mg) \times h_2 : (mg) \times h_3}$

$\large{\tt \leadsto W_1 : W_2 : W_3 = \cancel{(mg)} \times h_1 : \cancel{(mg)} \times h_2 : \cancel{(mg)} \times h_3}$

Therefore,

$\large{\tt \leadsto W_1 : W_2 : W_3 = h_1 : h_2 : h_3 \: ----(a)}$

$\large{\boxed{\tt W_1 : W_2 : W_3 = h_1 : h_2 : h_3 }}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

There is a constant acceleration on the ball I.e "g"(Acceleration due to gravity).

Now, Using nth second Formula,

$\large{\boxed{\tt S_n = u + \dfrac{a}{2}(2n - 1)}}$

Now, For Height h₁

$\large{\tt \leadsto h_1 = u + \dfrac{a}{2}(2n - 1)}$

Initial velocity (u) = 0 m/s.

$\large{\tt \leadsto h_1 = 0 + \dfrac{a}{2}(2 \times 1 - 1)}$

(Here n = 1 second because The body is falling uniformly)

$\large{\tt \leadsto h_1 = \dfrac{a}{2}(2 - 1)}$

$\large{\tt \leadsto h_1 = \dfrac{a}{2}[1] \: ----(4)}$

Now, For Height h₂

$\large{\tt \leadsto h_2 = u + \dfrac{a}{2}(2n - 1)}$

Initial velocity (u) = 0 m/s.

$\large{\tt \leadsto h_2 = 0 + \dfrac{a}{2}(2 \times 2 - 1)}$

(Here n = 2 second because The body is falling uniformly)

$\large{\tt \leadsto h_2 = \dfrac{a}{2}(4 - 1)}$

$\large{\tt \leadsto h_2 = \dfrac{a}{2}[3] \: ----(5)}$

Now, For Height h₃

$\large{\tt \leadsto h_3 = u + \dfrac{a}{2}(2n - 1)}$

Initial velocity (u) = 0 m/s.

$\large{\tt \leadsto h_3 = 0 + \dfrac{a}{2}(2 \times 3 - 1)}$

(Here n = 3 second because The body is falling uniformly)

$\large{\tt \leadsto h_3 = \dfrac{a}{2}(6 - 1)}$

$\large{\tt \leadsto h_3 = \dfrac{a}{2}[5] \: ----(6)}$

Now, Taking Ratio

$\large{\bold{\tt h_1 : h_2 : h_3}}$

Substituting the values,

$\large{\tt \leadsto h_1 : h_2 : h_3 = \dfrac{a}{2}[1] : \dfrac{a}{2}[3] : \dfrac{a}{2}[5]}$

$\large{\tt \leadsto h_1 : h_2 : h_3 = \cancel{\dfrac{a}{2}}[1] : \cancel{\dfrac{a}{2}}[3] : \cancel{\dfrac{a}{2}}[5]}$

it becomes,

$\large{\tt \leadsto h_1 : h_2 : h_3 = 1 : 3 :5 \: ----(b) }$

$\large{\boxed{\tt h_1 : h_2 : h_3 = 1 : 3 :5}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Substituting the value of equation (b) in equation (a)

$\large{\tt \leadsto W_1 : W_2 : W_3 = h_1 : h_2 : h_3 }$

$\large{\tt \leadsto W_1 : W_2 : W_3 = 1 : 3 : 5}$

$\huge{\boxed{\boxed{\tt W_1 : W_2 : W_3 = 1 : 3 : 5}}}$

Hence derived!

$\rule{300}{1.5}$