A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T
1 answer:
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)
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Answer:
X-Positions: Y-Positions
x(0) = 0 y(0) = 0
x(2) = 120 m y(2) = 19.6 m
x(4) = 240 m y(4) = 78.4 m
x(6) = 360 m y(6) = 176.4 m
x(8) = 480 m y(8) = 313 m
x(10) = 600m y (10) = 490 m
Explanation:
X-Positions
- First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
- After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
- So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:
- It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.
Y-Positions
- We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
- As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
- At any moment, it is subject to the acceleration of gravity, g.
- As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:
- Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
- y(2) = 2* 9.8 m/s2 = 19.6 m
- y(4) = 8* 9.8 m/s2 = 78.4 m
- y(6) = 18*9.8 m/s2 = 176.4 m
- y(8) = 32*9.8 m/s2 = 313.6 m
- y(10)= 50 * 9.8 m/s2 = 490.0 m
The formula for the rotational kinetic energy is
where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.
ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s
Then,
where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.
ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s
Then,