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Simora [160]
3 years ago
10

(a) At a certain instant, a particle-like object is acted on by a force F = (5.0 N) i hat - (2.0 N) j + (8.0 N) k while the obje

ct's velocity is v = - (2.0 m/s) i hat + (4.0 m/s) k. What is the instantaneous rate at which the force does work on the object?
Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

Power, P = 22 W

Explanation:

It is given that,

Force acting on the object, F=5i-2j+8k

Velocity of the object, v=-2i+4k  

To Find,

The rate at which the force does work on the object

Solve,

The rate at which the force does work on the object is called its power. Its formula is given by :

P=\dfrac{W}{t}

P=\dfrac{F.d}{t}

P=F\times v

P=(5i-2j+8k)\times (-2i+4k)

P=-10+0+32=22\ W

So, the force work on the object at the rate of 22 watts

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, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

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         k = F / x

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        x = A cos (wt + Ф)

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       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

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v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

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