Question

33. A particle starting from rest moves vertically upward from the surface of earth
under an external (variable) force F = (2 - az) mg k. His maximum height reached
by particle. (Vertical is along Z-axis)
4​

Answer 1

Given:

A particle starting from rest moves vertically upward from the surface of earth under an external (variable) force F = (2 - az).

To find:

Max height reached by the particle.

Calculation:

$\therefore \: F = 2 - az$

$= > m(acc.) = 2 - az$

$= > m \bigg \{v \times \dfrac{dv}{dz} \bigg \}= 2 - az$

$= > m \bigg \{v \times dv \bigg \}= (2 - az) \: dz$

Integrating on both sides:

$\displaystyle = > \int m \bigg \{v \times dv \bigg \}= \int(2 - az) \: dz$

$\displaystyle = > m\int v dv = \int(2 - az) \: dz$

Putting limits :

$\displaystyle = > m\int_{0}^{0} v dv = \int_{0}^{z} (2 - az) \: dz$

$= > \: 0 = 2z - \dfrac{a {z}^{2} }{2}$

$= > \: 0 = 2 - \dfrac{a z }{2}$

$= > \: \dfrac{a z }{2} = 2$

$= > \: az = 4$

$= > \: z = \dfrac{4}{a}$

So, final answer is :

$\boxed{ \sf{ \red{ \large{ \: z = \dfrac{4}{a} }}}}$

Answer 2

Explanation:

Given:

A particle starting from rest moves vertically upward from the surface of earth under an external (variable) force F = (2 - az).

To find:

Max height reached by the particle.

Calculation:

\therefore \: F = 2 - az∴F=2−az

= > m(acc.) = 2 - az=>m(acc.)=2−az

= > m \bigg \{v \times \dfrac{dv}{dz} \bigg \}= 2 - az=>m{v×

dz

dv

}=2−az

= > m \bigg \{v \times dv \bigg \}= (2 - az) \: dz=>m{v×dv}=(2−az)dz

Integrating on both sides:

\displaystyle = > \int m \bigg \{v \times dv \bigg \}= \int(2 - az) \: dz=>∫m{v×dv}=∫(2−az)dz

\displaystyle = > m\int v dv = \int(2 - az) \: dz=>m∫vdv=∫(2−az)dz

Putting limits :

\displaystyle = > m\int_{0}^{0} v dv = \int_{0}^{z} (2 - az) \: dz=>m∫

0

0

vdv=∫

0

z

(2−az)dz

= > \: 0 = 2z - \dfrac{a {z}^{2} }{2}=>0=2z−

2

az

2

= > \: 0 = 2 - \dfrac{a z }{2}=>0=2−

2

az

= > \: \dfrac{a z }{2} = 2=>

2

az

=2

= > \: az = 4=>az=4

= > \: z = \dfrac{4}{a}=>z=

a

4

So, final answer is :

\boxed{ \sf{ \red{ \large{ \: z = \dfrac{4}{a} }}}}

z=

a

4

.....