When you set a heavy bag down on the ground, you are doing _______ work on it.
1 answer:
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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)
Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass,
Area of cross section,
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
Weight of the object = drag force
R = W
or
Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,
So, the drag coefficient is 1.01.