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saw5 [17]
3 years ago
7

16. A student uses 110 N of force to push a box

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0

Answer:

792J

Explanation:

Use the formula Force × Distance

With 110 newtons being the force and 7.2 meters being the distance

Which gives 792 and work us measured in joules therefore its joules

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The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di
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Answer and Explanation:

Let:

x(t)=Acos(\omega t+ \phi)

The equation representing a simple harmonic motion, where:

x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)

Also, you may know these fundamental formulas:

f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m

(b)

v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s

(c)

a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

\phi = \frac{\pi}{5}

(e)

f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz

(f)

T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s

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