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Solnce55 [7]
2 years ago
5

A free undamped spring/mass system oscillates with a period of 4 seconds. When 10 pounds are removed from the spring, the system

then has a period of 2 seconds. What was the weight of the original mass on the spring? (Round your answer to one decimal place.)
Physics
1 answer:
OverLord2011 [107]2 years ago
6 0

Answer:

13.3 pounds.

Explanation:

For a spring of constant K, with an attached object of mass M, the period can be written as:

T = 2*π*√(M/K)

Where π = 3.14

First, we know that the period is 4 seconds, then we have:

4s = (2*π)*√(M/K)

We know that if the mass is reduced by 10lb, the period becomes 2s.

Then the new mass of the object will be: (M - 10lb)

Then the period equation becomes:

2s = (2*π)*√((M-10lb)/K)

So we have two equations:

4s = (2*π)*√(M/K)

2s = (2*π)*√((M-10lb)/K)

We want to solve this for M.

First, we need to isolate K in one of the equations.

Let's isolate K in the first one:

4s = (2*π)*√(M/K)

(4s/2*π) = √(M/K)

(2s/π)^2 = M/K

K = M/(2s/π)^2 = M*(π/2s)^2

Now we can replace it in the other equation.

2s = (2*π)*√((M-10lb)/K)

First, let's simplify the equation:

2s/(2*π) = √((M-10lb)/K)

1s/π =  √((M-10lb)/K)

(1s/π)^2 =  ((M-10lb)/K

K*(1s/π)^2 = M - 10lb

Now we can use the equation: K =  M*(π/2s)^2

then we get:

K*(1s/π)^2 = M - 10lb

(M*(π/2s)^2)*(1s/π)^2 = M - 10lb

M/4 = M - 10lb

10lb = M - M/4

10lb = (3/4)*M

10lb*(4/3) = M

13.3 lb = M

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Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

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3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

5.3 m W = - 5.3 i

3.7 m N, = 3.7 j

5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = -  4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

= -4.6 i -7.1 j

magnitude of displacement = \sqrt{(4.6^2+7.1^2)}

= 8.46 m

Direction

Tanθ = 7.1/ 4.6

θ = 57⁰ south of west .

distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9

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3 years ago
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A box with mass (m) it's sliding along on a friction-free surface at 9.87 m/s at a height of 1.81 meters. It travels down the hi
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A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² =  1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² =  1/2 vi² + a * h

We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m

1/2 vf² =  1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s


b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
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v = 11.53 m/s
a = 9.8 m/s² 
h = ?

1/2 * 11.53² = 9.8 * h
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66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
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Answer:

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D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

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θ =11.25 ° south of due west.

4 0
3 years ago
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