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nadezda [96]
2 years ago
12

An ideal spring obeys hooke's law: f = −bikx. a mass of m = 0.3 kg hung vertically from this spring stretches the spring 0.13 m.

the acceleration of gravity is 9.8 m/s 2 . calculate the value of the force constant k for the spring. answer in units of n/m
Physics
1 answer:
Anarel [89]2 years ago
5 0
The force = 0.3 * 9.8 = 2.94
x = 0.13

k = 2.94/0.13
k = 22.6 N/m
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1.) A stone falls from rest from the top of a cliff.
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Explanation:

ignore air resistance

Let t be the time of fall for the dropped stone.

½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²

4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)

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h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m

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Vocabulary word for the total amount of charge in a closed system remains constant
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In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom
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Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2  (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R =  38.5 x ( v^2 )

77 x 9.8 x 3.65 =  38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) =  71.54

v = 8.4 m/s  ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)  

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf  

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy  = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0
3 years ago
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