Answer:
F = 0.408 N
Explanation:
It is given that,
Weight of a wooden block, W = 8 N
Weight, W = mg
m is mass of wooden block
![m=\dfrac{W}{g}\\\\m=\dfrac{8\ N}{9.8\ m/s^2}\\\\m=0.816\ kg](https://tex.z-dn.net/?f=m%3D%5Cdfrac%7BW%7D%7Bg%7D%5C%5C%5C%5Cm%3D%5Cdfrac%7B8%5C%20N%7D%7B9.8%5C%20m%2Fs%5E2%7D%5C%5C%5C%5Cm%3D0.816%5C%20kg)
Acceleration of the block, a = 0.5 m/s²
Force, F = ma
![F=0.816\ kg\times 0.5\ m/s^2\\\\F=0.408\ N](https://tex.z-dn.net/?f=F%3D0.816%5C%20kg%5Ctimes%200.5%5C%20m%2Fs%5E2%5C%5C%5C%5CF%3D0.408%5C%20N)
So, the magnitude of force applied to the wooden block is 0.408 N.
![\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Cboxed%7B%20%5Cbf%7B%20%5Ccolor%7Bred%7D%7BUniversal%20%5C%3A%20law%20%5C%3A%20of%20%5C%3A%20gravitation%7D%7D%7D%7D)
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.
❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.
According to universal law of gravitation,
![\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Clongrightarrow%7B%20%5Crm%7BF%20%5Cpropto%20m_1%20m_2%7D%7D%7D)
Also,
![\large{ \longrightarrow{ \rm{ F \propto \dfrac{1}{ {d}^{2} } }}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Clongrightarrow%7B%20%5Crm%7B%20F%20%5Cpropto%20%20%5Cdfrac%7B1%7D%7B%20%7Bd%7D%5E%7B2%7D%20%7D%20%7D%7D%7D)
Combining both, We will get
![\large{ \longrightarrow{ \rm{F \propto \dfrac{ m_1 m_2}{ {d}^{2}}}}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Clongrightarrow%7B%20%5Crm%7BF%20%20%5Cpropto%20%20%5Cdfrac%7B%20m_1%20m_2%7D%7B%20%7Bd%7D%5E%7B2%7D%7D%7D%7D%7D%20)
Or, We can write it as,
![\large{ \longrightarrow{ \rm{F \propto \: G \dfrac{ m_1 m_2}{ {d}^{2} }}}}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Clongrightarrow%7B%20%5Crm%7BF%20%20%5Cpropto%20%20%5C%3A%20%20G%20%5Cdfrac%7B%20m_1%20m_2%7D%7B%20%7Bd%7D%5E%7B2%7D%20%7D%7D%7D%7D)
Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.
☯️ Hence, derived !!
<u>━━━━━━━━━━━━━━━━━━━━</u>
On the half way down (object falling down with the initial speed = 0)
Answer:
14.4 m/s
Explanation:
= initial velocity of the projectile at the time of launch = 25 m/s
= initial angle of launch = 60°
= initial velocity of the projectile in horizontal direction=
= 25 Cos60 = 12.5 m/s
= final velocity of the projectile = ![v](https://tex.z-dn.net/?f=v)
= final angle = 30°
= final velocity of the projectile in horizontal direction=
=
Cos30
we know that the velocity along the horizontal direction remains constant, hence
=
Cos30 = 12.5
= 14.4 m/s