Question

A body of mass 10 kg is kept at a height of 5 m. It is allowed to fall and reach the ground.

(i) What is the total mechanical energy possessed by the body at the height of 2 m assuming it is a frictionless medium ?

(ii) What is the kinetic energy possessed by the body just before hitting the ground ?
(Take g = 10 m/s².)​

Answer 1

$\huge\boxed{\fcolorbox{aqua}{lime}{answer☆}}$

(a) (i) The total mechanical energy is the total potential energy it possesses before its fall.

E = mgh = 10 × 10 × 5 = 500 J 

It is equal to the maximum potential energy E = mgh = 10 × 10 × 5 = 500 J. 

(b) (i) The weight of the metre scale acts at its centre of gravity i.e.,

at 50 cm mark.

By the concept of moments, we have  40 × (30 – 5 ) = W × (50 – 30)  or W = (40 × 25)/ 20 = 50 gf 

(ii) F is shifted towards 0 cm. 

(c) (i) The direction of tension is as marked.

(ii) VR = Distance travelled by effort/

Distance travelled by load = 2x/2= 2

. The distance travelled by load is half the distance moved by effort =x/2 

(iii) T = E. 

(iv) The distance travelled by load is half the distance moved by effort = x/2.

Answer 2

★ Concept :-

Here the concept of Kinetic Energy and Potential Energy has been used. We see that we need to calculate the mechanical energy and kinetic energy of the body in both cases. So firstly, we shall find the velocity of the body in both cases. Then we shall find potential energy of the body and then kinetic energy. And thus we can derive our further answer

Let's do it !!. ______________________________________

★ Formula Used :-

$\;\boxed{\sf{\pink{Potential\;Energy\;=\;\bf{m\:g\:h}}}}$

$\;\boxed{\sf{\pink{v^{2}\;-\;u^{2}\;=\;\bf{2as}}}}$

$\;\boxed{\sf{\pink{Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:m\:v^{2}}}}}$

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★ Solution :-

Given,

» Mass of the body = m = 10 Kg

» Initial height of body = 5 m

______________________________________

i.) Given,

• Initial velocity of body = u = 0 m/sec

• Final velocity of body = v

• Distance covered = s = 5 - 2 = 3 m

(since body covers this displacement after falling)

• New height of body = h = 2 m

• Acceleration due to gravity = a = 10 m/sec²

From Third Equation of Motion we get,

$\;\sf{\rightarrow\;\;v^{2}\;-\;u^{2}\;=\;\bf{2as}}$

By applying values, we get

$\;\sf{\rightarrow\;\;v^{2}\;-\;(0)^{2}\;=\;\bf{2(10)(3)}}$

$\;\bf{\rightarrow\;\;\green{v^{2}\;=\;\bf{60\;m^{2}\:sec^{-2}}}}$

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From the formula of Potential Energy, we get

$\;\sf{\rightarrow\;\;Potential\;Energy\;=\;\bf{m\:g\:h}}$

By applying values, we get

$\;\sf{\rightarrow\;\;Potential\;Energy\;=\;\bf{(10)(10)(2)}}$

$\;\bf{\rightarrow\;\;Potential\;Energy\;=\;\bf{\orange{200\;\:J}}}$

From the formula of Kinetic Energy, we get

$\;\bf{\rightarrow\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:m\:v^{2}}}$

By applying values, we get

$\;\sf{\rightarrow\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:\times\:10\:\times\:60}}$

$\;\sf{\rightarrow\;\;Kinetic\;Energy\;=\;\bf{5\:\times\:60}}$

$\;\bf{\rightarrow\;\;Kinetic\;Energy\;=\;\bf{\blue{300\;\;J}}}$

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We know that,

✒ Total Mechanical Energy = Kinetic Energy + Potential Energy

✒ Total Mechanical Energy = 200 J + 300 J

✒ Total Mechanical Energy = 500 J

$\;\underline{\boxed{\tt{Total\;\:mechanical\;\:energy\;=\;\bf{\purple{500\;\;J}}}}}$

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ii.) Given,

» Initial Velocity of body = u = 0 m/sec

» Final Velocity of body = v

» Displacement covered = s = 5 m

(since body falls down and we need to calculate kinetic energy just before striking ground)

» Acceleration due to gravity = a = 10 m/sec²

From the Third Equation of Motion, we get

$\;\sf{\rightarrow\;\;v^{2}\;-\;u^{2}\;=\;\bf{2as}}$

By applying values, we get

$\;\sf{\rightarrow\;\;v^{2}\;-\;(0)^{2}\;=\;\bf{2(10)(5)}}$

$\;\bf{\rightarrow\;\;\green{v^{2}\;=\;\bf{100\;m^{2}\:sec^{-2}}}}$

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From formula of the Kinetic Energy, we get

$\;\bf{\rightarrow\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:m\:v^{2}}}$

By applying the values, we get

$\;\sf{\rightarrow\;\;Kinetic\;Energy\;=\;\bf{\dfrac{1}{2}\:\times\:10\:\times\:100}}$

$\;\sf{\rightarrow\;\;Kinetic\;Energy\;=\;\bf{5\:\times\:100}}$

$\;\bf{\rightarrow\;\;Kinetic\;Energy\;=\;\bf{\blue{500\;\;J}}}$

$\;\underline{\boxed{\tt{Kinetic\;\:energy\;=\;\bf{\purple{500\;\;J}}}}}$

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★ More to know :-

$\;\sf{\leadsto\;\;W\;=\;F\:\times\:s}$

$\;\sf{\leadsto\;\;F\;=\;ma}$

$\;\sf{\leadsto\;\;P\;=\;\dfrac{Workdone}{Time}}$