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sukhopar [10]
3 years ago
13

The gravitational acceleration on the surface of the moon is 1/6 what it is on earth. If a man could jump straight up 0.7 m (abo

ut 2 feet) on the earth, how high could he jump on the moon?
Physics
1 answer:
LenaWriter [7]3 years ago
5 0

Answer:

on moon he can jump 4.2 m high

Explanation:

given data

gravitational acceleration on moon a(m) = 1/6

jump = 0.7 m

to find out

how high could he jump on the moon

solution

we know gravitational acceleration on earth a =  g = 9.8 m/s²

so on moon am =  9.8 * \frac{1}{6}  = 1.633 m/s²

so if he jump on earth his speed will be for height 0.7 m i s

speed v = \sqrt{2gh}

v = \sqrt{2(9.8)0.7}

v = 3.7 m/s

so if he hump on moon

height will be

height = \frac{v^2}{2*a(m)}  

put here value

height =  \frac{3.7^2}{2*1.633)}  

height = 4.2 m

so on moon he can jump 4.2 m high

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3 years ago
A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The
Yuki888 [10]

Answer:

Approximately 0.560\; {\rm m}, assuming that:

  • the height of 3.21\; {\rm m} refers to the distance between the clay and the top of the uncompressed spring.
  • air resistance on the clay sphere is negligible,
  • the gravitational field strength is g = 9.81\; {\rm m\cdot s^{-2}}, and
  • the clay sphere did not deform.

Explanation:

Notations:

  • Let k denote the spring constant of the spring.
  • Let m denote the mass of the clay sphere.
  • Let v denote the initial speed of the spring.
  • Let g denote the gravitational field strength.
  • Let h denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let x denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of x, the elastic potential energy \text{PE}_{\text{spring}} in this spring would be:

\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}.

The initial kinetic energy \text{KE} of the clay sphere was:

\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}.

When the spring is at the maximum compression:

  • The clay sphere would be right on top of the spring.
  • The top of the spring would be below the original position (when the spring was uncompressed) by x.
  • The initial position of the clay sphere, however, is above the original position of the top of the spring by h = 3.21\; {\rm m}.

Thus, the initial position of the clay sphere (h = 3.21\; {\rm m} above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by (h + x).

The gravitational potential energy involved would be:

\text{GPE} = m\, g\, (h + x).

No mechanical energy would be lost under the assumptions listed above. Thus:

\text{PE}_\text{spring} = \text{KE} + \text{GPE}.

\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x).

Rearrange this equation to obtain a quadratic equation about the only unknown, x:

\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0.

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(The other root is negative and is thus invalid.)

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