Question

A ball is released from the top of a tower. The ratio of work done by force of gravity in $1^{st}$ second, $2^{nd}$ second and $3^{rd}$ second of the motion of ball is

• A. $1:2:3$
• B. $1:4:16$
• C. $1:3:5$
• D. $1:9:25$

Answer 1

The correct option is C $1:3:5$

As the ball is released, initial velocity of ball is zero i.e., $u=0$

We know that displacement in $t$ seconds : $s=ut+\frac{1}{2}a{t}^2$

Displacement for the $1^{st}$ second is : $s_1=\frac{1}{2}\times 10\times 1=5$

Displacement for the $2^{nd}$ second is: displacement for $2\text{s}-$ displacement for $1\text{s}$
$s_2=\frac{1}{2}\times 10\times (2{)}^2-\frac{1}{2}\times 10\times (1{)}^2$
$s_2=20-5=15$

Displacement for the $3^{rd}$ second is: displacement for $3\text{s}-$ displacement for $2\text{s}$
$s_3=\frac{1}{2}\times 10\times (3{)}^2-\frac{1}{2}\times 10\times (2{)}^2$
$s_3=45-20=25$

or $s_1:s_2:s_3=1:3:5$

Now, $W=mgs$
$\Rightarrow W\propto s$
$\therefore W_1:W_2:W_3=1:3:5$

Hence option C is the correct answer