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4 years ago

8

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find

the initial speed of the rock.
a. 3m/s
b. 30.3 m/s
c. None of the above

1 answer:

butalik [34]4 years ago

5 0

Answer:

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy ..............................1

so it can be written as

work = force × distance ...................2

and

KE is express as

K.E = 0.5 × m × v²

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)² ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² = 919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

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