Question

о со особо охоор
10. A 1 kg mass falls from a height of 10 m into a large box filled with sand. What is the speed of the mass just
before hitting the box? What is its kinetic energy?
The mass travels 2 cm into the sand before coming to rest. What is the average retarding force of sand?
[Ars: 14 m S-1, 98 J, 4900 N]​

Answer 1

Answer:

Explanation:

Hey, here's your answer attached.

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Height (h) = 10m.
• Mass of body(m) = 1 kg .
• Distance travelled in sand = 2 cm = 0.02 cm.

$\huge{\bold{\underline{Explanation:-}}}$

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By Conservation of energy,

$\large{\boxed{ \text{ Increase in Kinetic energy = Decrease in Potential energy}}}$

I.e.

$\large{\tt \leadsto \uparrow Kinetic \: energy = \downarrow Potential \: energy}$

Substituting the values,

$\tt \leadsto \dfrac{1}{2}mv^2 = mgh \\ \tt \leadsto \dfrac{1}{2}\cancel{m}v^2 = \cancel{m}gh \\ \tt \leadsto \dfrac{1}{2}.v^2 = gh \\ \tt \leadsto v^2 = 2gh \\ \tt \leadsto v = \sqrt{2gh} \\ \text{Substituting the values} \\ \tt \leadsto v = \sqrt{2 \times 10 \times 10} \\ \tt \leadsto v = \sqrt{200} \\ \tt \leadsto v = 10 \sqrt{2} \\ \text{As we know value i.e. 1.414} \\ \tt \leadsto v = 10 \times 1.414$

$\huge{\boxed{\boxed{\tt v = 14.14 \: m/s}}}$

So, the velocity of the body is 14.14 m/s.

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From The Kinetic energy formula,

$\large{\boxed{\tt K.E = \dfrac{1}{2}mv^2}}$

Substituting the values ,

$\tt \leadsto K.E = \dfrac{1}{2}mv^2 \\ \tt \leadsto K.E = \dfrac{1}{2} \times 1 \times (14.1)^2 \\ \text{The velocity is taken from the above relation} \\ \tt \leadsto K.E = \dfrac{1}{2} \times 1 \times 196 \\ \tt \leadsto K.E = \dfrac{1}{\cancel{2}} \times 1 \times \cancel{196} \\ \tt \leadsto K.E = 1 \times 98 \:$

$\huge{\boxed{\boxed{\tt K.E = 98 \: J}}}$

So, the Kinetic energy of the body is 98 Joules.

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Now, By work - Energy theorem,

$\large{\boxed{\tt W = \dfrac{1}{2}mv^2}}$

Substituting the values,

$\tt \leadsto F.s = \dfrac{1}{2} mv^2 \\ \underline{\text{As Work done = F.s}} \\ \text{Substituting the values} \\ \tt \leadsto F \times 0.02 = \dfrac{1}{2} \times 1 \times (14)^2 \\ \tt \leadsto F \times 0.02 = 98 \\ \tt \leadsto F = \dfrac{98}{0.02} \\ \\ \tt \leadsto F = \dfrac{98 \times 100}{2} \\ \tt \leadsto F = \dfrac{\cancel{98} \times 100}{\cancel{2}} \\ \tt \leadsto F = 49 \times 100 \:$

$\huge{\boxed{\boxed{\tt F = 4900 \: N}}}$

So, the Force acting on the body is 4900N.

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