Question

A ball thrown vertically upwards with a speed of 19.6 m/s from the top of a tower return to the earth in 6 s. What is the height of the tower?
state and prove work-Energy Theorem

Answer 1

Given that,

Initial velocity $u=19.6m/s$

Final velocity $v=0$

Time $t=6sec$

The acceleration is

We know that,

$v=u+at$

$0=19.6+a\times 6$

$a=-3.26m/s^2$

Now, the height is

From equation of motion

$s=ut+\frac{1}{2}a{t}^2$

$s=19.6\times 6-\frac{1}{2}\times 3.26\times 36$

$s=58.9$

$s=59m$

Hence, the height of the tower is $59m$

The work–energy theorem: -

The work done by all forces on an object is equal to the change in kinetic energy of the object.

$W=\Delta k$

From newton second law,

$F=ma....\left(I\right)$

We know that,

$a=\frac{dv}{dt}$

So, put the value of a in equation (I)

$F=m\frac{dv}{dt}$

If multiple both side by $v$

$F.v=mv\frac{dv}{dt}$.....(II)

We know that,

$v=\frac{dx}{dt}$

Now, put the value of v in equation (II)

$F\frac{dx}{dt}=mv\frac{dv}{dt}$

Now, on integrating

$Fdx=mvdv$

$F\underset{0}{\overset{x}{\int }}dx=m\underset{v_1}{\overset{v_2}{\int }}vdv$

$Fx=\frac{1}{2}m(v_2^2-v_1^2)$

$Fx=\frac{1}{2}m{v}_2^2-\frac{1}{2}m{v}_1^2$

We know that,

$W=Fx$

$\frac{1}{2}m{v}^2=k$

$k$ = kinetic energy

Now,

$W=k_2-k_1$

$W=\Delta k$

The work –energy theorem is proved.