Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.
1 answer:
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Answer:
F=1.65 x 10²⁶ N
Explanation:
Given that
Distance ,R= 3.34 x 10¹² m
Mass m₁= 2.78 x 10³⁰ kg
Mass ,m₂= 9.94 x 10³⁰ kg
we know that gravitational force F given as
G=Constant
G=6.67 x 10⁻¹¹ Nm²/kg²
Now by putting the values
F=1.65 x 10²⁶ N
Therefore the force between these two mass will be 1.65 x 10²⁶ N.
Answer:
a)5.88J
b)-5.88J
c)0.78m
d)0.24m
Explanation:
a) W by the block on spring is given by
W= kx² =
(530)(0.149)² = 5.88 J
b) Workdone by the spring = - Workdone by the block = -5.88J
c) Taking x = 0 at the contact point we have U top = U bottom
So, mg =
kx² - mgx
And, = (
kx² - mgx
)/(mg) =
]/(0.645x9.8)
= 0.78m
d) Now, if the initial initial height of block is 3
= 3 x 0.78 = 2.34m
then, kx² - mgx - mg
=0
(530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0
265x² - 6.321x - 14.8 = 0
a=265
b=-6.321
c=-14.8
By using quadratic eq. formula, we'll have the roots
x= 0.24 or x=-0.225
Considering only positive root:
x= 0.24m (maximum compression of the spring)