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yKpoI14uk [10]
2 years ago
8

A 5 kg object is moving in a straight-line with an initial speed of v m/s. It takes 13 s for the speed of the object to increase

to 13 m/s and it kinetic energy increases at a rate of 15 J/s. What is the initial speed v (in m/s)?
Physics
1 answer:
coldgirl [10]2 years ago
8 0

The object's kinetic energy changes according to

d<em>K</em>/d<em>t</em> = 15 J/s

If <em>v</em> is the object's initial speed, then its initial kinetic energy is

<em>K</em> (0) = 1/2 (5 kg) <em>v</em> ²

Use the fundamental theorem of calculus to solve for <em>K</em> as a function of time <em>t</em> :

K(t) = K(0) + \displaystyle\int_0^t \left(15\frac{\rm J}{\rm s}\right)\,\mathrm du = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)t

After <em>t</em> = 13 s, the object's kinetic energy is

<em>K</em> (13 s) = 1/2 (5 kg) (13 m/s)² = 422.5 J

Put this as the left side in the equation above for <em>K(t)</em> and solve for <em>v</em> :

422.5\,\mathrm J = \dfrac12 (5\,\mathrm{kg}) v^2 + \left(15\dfrac{\rm J}{\rm s}\right)(13\,\mathrm s)

==>   <em>v</em> ≈ 9.5 m/s

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r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

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time, t = 35 s

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using equation of motion

v = u + a t

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v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

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h₂ = 76562.5 m

total height = h₁ + h₂

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now, time taken by before the rocket hit the ground

using equation of motion

s = u t +\dfrac{1}{2}at^2

-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}

t = 260.57 s

neglecting the negative sign

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T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

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jasenka [17]

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The can be found elsewhere and as follows:


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