hlo anyone help
a stone is thrown upward with the kinetic energy of 10 joule if it goes up to a maximum height of 5 M find the initial velocity and mass of the stone??
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Answers

Answer 1

Explanation:

Here, kinetic energy of the body, K.E=10 J

Height attained by the body, h=5 m

h=5m

When the body attains maximum height, its kinetic energy is converted into potential energy.

P.E=K.E

⟹mgh=10

⟹m= 10/gh

= m = 10/(10×5) =0.2 kg (taking the value g as 10m/s²)

Now, ½mv²=10

0.2×v²=20

v²=20/0.2

v²=100

v=√100

v=10m/s.

hope this helps you.

Answer 2

[tex]\huge \bf༆ Answer ༄[/tex]

Let's solve ~

As we know the total energy [P.E + K.E = C] of the system remains constant throughout the motion,

So, When a stone was thrown initially it didn't had any potential energy (P.E = 0) but had kinetic Energy of 10 joules.

So, total energy = P.E + K.E = 0 + 10 = 10 joules

As per the given information, equate it with the formula.

[tex] \sf \dfrac{1}{2}m {v}^{2} = 10[/tex]

[tex] \sf m {v}^{2} = 20[/tex]

Now, As it approaches 5m height it comes to rest, Therefore velocity = 0. And since velocity = 0 then Kinetic Energy = 0

Now, let's find the Potential Energy at that point ~

[tex] \sf mgh[/tex]

[tex] \sf m \times 10 \times 5[/tex]

[tex] \sf50m[/tex]

And here, the Total energy = P.E + K.E = 10 Joules

So,

[tex] \sf50m + 0 = 10 [/tex]

[tex] \sf50m = 10[/tex]

[tex] \sf m = 10 \div 50[/tex]

[tex] \sf m = 0.2[/tex]

Therefore, mass of the object is 0.2 kg = 200 grams

Now, plug the value of mass (m) in the equation of kinetic Energy to find the initial velocity of the stone ~

[tex] \sf m {v}^{2} = 20[/tex]

[tex] \sf0.2 \times {v}^{2} = 20[/tex]

[tex] \sf {v}^{2} = 20 \div 0.2[/tex]

[tex] \sf v = \sqrt{100} [/tex]

[tex] \sf v = 10 \: \: ms {}^{ - 1} [/tex]

Hence, velocity of the particle at the beginning was 10 m/s


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Hi there!

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hope this helped

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