Question

Physics Questions...

1) Find the acceleration due to gravity(g) on the surface of the earth given that radius of the earth is 6.37 × 10³ km, mass of the earth is 5.98×10²⁴kg.

2) A body can apply a force of 20 Newton and pull a box at constant speed of 2 m/s. Calculate power?

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Answer 1

${\large{\pmb{\sf{\underline{QuEsion \: 1^{st}}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Given \: that...}}}}}$

⋆ Mass of the body = 5.98 × 10²⁴ kilograms.

⋆ Radius of the body = 6.37 × 10³ kilometres.

⋆ The value of acceleration due to gravity that is g = ?

• g = 9.87 m/s²

${\bigstar \:{\pmb{\sf{\underline{Using \: concepts...}}}}}$

⋆ The value of gravitational constant.

⋆ Formula to convert kilometres into metres.

⋆ Some law of exponents.

⋆ Formula to find out the value of g (acceleration due to the gravity)

${\bigstar \:{\pmb{\sf{\underline{Using \: formulas...}}}}}$

${\small{\underline{\boxed{\sf{\star \: 1 \: kilometres \: = 1000 \: metres}}}}}$

${\small{\underline{\boxed{\sf{\star \: Value \: of \: G \: = 6.7 \times 10^{-11} \: Nm^2 kg^{-2}}}}}}$

(Where, G denotes gravitational constant)

Using law of exponents:

$\quad \quad{\sf{\bull a^b \times a^c = a^{b+c}}}$

$\quad \quad{\sf{\bull \dfrac{?}{a^{-c}} \: = \: ? \times a^c}}$

(Here, this law means if any exponent have negative power and it is in denominator then when moved to numerator the exponent power be in positive. Vice versa for its inverse)

${\small{\underline{\boxed{\sf{\star \: g \: = G\dfrac{M}{R^2}}}}}}$

(Where, g denotes acceleration due to gravity , G denotes gravitational constant , Mass denotes mass of object , R denotes radius)

${\bigstar \:{\pmb{\sf{\underline{Full \: Solution...}}}}}$

~ Firstly let us change kilometres into metres by using the suitable formula!

$:\implies \sf 1 \: kilometres \: = 1000 \: metres \\ \\ :\implies \sf 6.37 \times 10^3 = 6.37 \times 10^3 \times 1000 \\ \\ :\implies \sf 6370000 \: metres \\ \\ \tt In \: exponential \: form \\ \\ :\implies \sf 6.37 \times 10^6 \: metres$

~ Now let's use the formula to find out the value of acceleration due to gravity (g) by using suitable formula. And there let us imply suitable values.

$:\implies \sf g \: = G\dfrac{M}{R^2} \\ \\ :\implies \sf g = \dfrac{6.7 \times 10^{-11} \times 5.98 \times 10^{24}}{(6.37 \times 10^6)^{2}} \\ \\ :\implies \sf g = \dfrac{6.7 \times 10^{-11} \times 5.98 \times 10^{24}}{6.37 \times 10^6 \times 6.37 \times 10^6} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{-11} \times 10^{24} \times 10^{-6} \times 10^{-6}}{6.37 \times 6.37} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{-11} \times 10^{-6} \times 10^{-6} \times 10^{24}}{6.37 \times 6.37} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{-11 + (-6) + (-6)} \times 10^{24}}{6.37 \times 6.37} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{-11-6-6} \times 10^{24}}{6.37 \times 6.37} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{-17-6} \times 10^{24}}{6.37 \times 6.37} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{-23} \times 10^{24}}{6.37 \times 6.37} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{-23+24}}{6.37 \times 6.37} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{1}}{6.37 \times 6.37} \\ \\ :\implies \sf g = \dfrac{6.7 \times 5.98 \times 10^{1}}{40.5769} \\ \\ :\implies \sf g = \dfrac{400.66}{40.5769} \\ \\ :\implies \sf g = \dfrac{400660000}{40576900} \\ \\ :\implies \sf g = \dfrac{4006600}{405769} \\ \\ :\implies \sf g = 9.87 \: m/s$

____________________

${\large{\pmb{\sf{\underline{QuEsion \: 2^{nd}}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Understanding \: the \: question...}}}}}$

⋆ This question says that we have to find out the power when a body can apply a force of 20 Newton and pull a box at constant speed of 2 m/s.

${\bigstar \:{\pmb{\sf{\underline{Provided \: that...}}}}}$

⋆ Force = 20N (Given)

⋆ Constant speed = 2 m/s (Given)

⋆ Power = ? (To find)

• Power = 40 Watts

${\bigstar \:{\pmb{\sf{\underline{Full \: Solution...}}}}}$

~ As we know that

${\small{\underline{\boxed{\sf{\star \: Power \: = \dfrac{Work \: done}{Time}}}}}}$

~ We also know that

${\small{\underline{\boxed{\sf{\star \: Work \: done \: = Force \times Displacement \: or \: Distance}}}}}$

~ We also know that

${\small{\underline{\boxed{\sf{\star \: Average \: speed \: = \dfrac{Total \: distance}{Total \: time}}}}}}$

• Therefore,

${\small{\underline{\boxed{\sf{\star \: v \: = \dfrac{s}{t}}}}}}$

~ We can write this as

${\small{\underline{\boxed{\sf{\star \: P \: = \dfrac{f \times s}{t}}}}}}$

${\small{\underline{\boxed{\sf{\star \: P \: = fv}}}}}$

~ So according to condition

$:\implies \sf P \: = fv \\ \\ :\implies \sf P \: = 20(2) \\ \\ :\implies \sf P \: = 20 \times 2 \\ \\ :\implies \sf P \: = 40 \: W \\ \\ :\implies \sf Power \: = 40 \: Watts$