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lesantik [10]
3 years ago
5

A force =(16.8N/s)t is applied to an object of m=45.0kg. Ignoring friction, how far does the object travel from rest in 5.00s?

Physics
1 answer:
kirza4 [7]3 years ago
6 0

Answer:

Explanation:

a = F/m = 16.8t/45 = 0.373*t

v = ∫a*dt = ∫(0.373*t)dt = 0.1865*t²

x = ∫v*dt = 0.1865∫(t²)dt = 0.062*t³

For t = 5.0 sec,

x = 0.062*(5)^3 = 7.75 m

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A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
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Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

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E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

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