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Lerok
12 days ago
9

You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i

s placed against a light spring that is compressed 0.280 m. The spring has force constant 48.0 N/m . The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is μk = 0.300. When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. (a) What is the speed of the box at the instant when it leaves the spring? (b) What is the maximum speed of the box during its motion?
Physics
1 answer:
ValentinkaMS [1.1K]12 days ago
6 0

Answer:

a) V = 0.82m/s

b) Vmax = 0.985 m/s

Explanation:

According to the conservation of energy, we know:

Eo = Ef    \frac{m*V^{2}}{2}-\frac{K*Xmax^{2}}{2}=-Ff*Xmax

We find V by solving this:

V = 0.82 m/s

To calculate the maximum speed, we will also solve for an intermediate position where compression is X, and the work done by friction is given as (Xmax - X) = (0.28m - X):

\frac{m*Vmax^{2}}{2}+\frac{K*X^{2}}{2}-\frac{K*Xmax^{2}}{2}=-Ff*(Xmax-X)

We will then differentiate and set the result to zero to find position X. After performing the derivative, we discover:

X = 0.1m    Putting this back into the equation for Vmax:

Vmax = 0.985m/s

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kicyunya [1025]

To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:

r = 16m

m = 82kg

\mu_s = 0.63

The maximum velocity can be determined using centripetal force,

F_c = \frac{mv^2}{r}

Should be equal to,

\frac{mv^2}{r} = \mu_s mg

v = \sqrt{\mu_s gr}

v = \sqrt{(0.63)(9.8)(16)}

v = 9.93m/s

As a result, the highest speed achievable through the arc without slipping is 9.93m/s

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13 hours ago
A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
serg [1198]
The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.
8 0
4 days ago
Jack pulls a sled across a level field by exerting a force of 110 n at an angle of 30 with the ground. what are the parallel and
Softa [913]
<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.

For the component parallel to the ground:
x = rcos</span>β
<span>x = 110cos30</span>°
<span>x = 95.26

For the component perpendicular to the ground:
y = rsin</span>β
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3 0
6 days ago
Read 2 more answers
A champion athlete can produce one horsepower (746 W) for a short period of time. The number of 16-cm-high steps a 70-kg athlete
Sav [1105]

Answer:

407 steps

Explanation:

Based on the question,

P = mgh/t........... Equation 1

Where P stands for power, m is mass, g denotes gravity, h is height, and t represents time.

Rearranging the equation to solve for h, we have:

h = Pt/mg............. Equation 2

Providing values: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.

Given constant: g = 9.8 m/s²

By substituting into equation 2

h = 746(60)/(70×9.8)

h = 44760/686

h = 65.25 m

h = 6525 cm

Calculating number of steps: 6525/16

The resulting number of steps = 407 steps

6 0
9 days ago
What happens to the particles of a liquid when energy is removed from them?
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Response:

D: The distance among the particles diminishes

Clarification:

Removing energy reduces the activity of molecules, similar to how one slows down in cold temperatures (I believe).

3 0
10 days ago
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