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Alex_Xolod [135]
3 years ago
11

For a moving object, the force acting on the object varies directly with the objects acceleration. When a force of 24N acts on a

certain object, the acceleration of the object is 3m/s^2. If the acceleration of the object becomes 10m/s^2, what is the force
Physics
1 answer:
VikaD [51]3 years ago
6 0

Answer:

80 N

Explanation:

From the question,

F = ma .......................... Equation 1

Where F = Force, m = mass of the object, a = acceleration of the object.

make m the subject of the equation

m = F/a.................... Equation 2

Given: F = 24 N, a = 3 m/s²

Substitute into equation 2

m = 24/3

m = 8 kg.

If the acceleration of the object becomes 10 m/s²,

Substitute into equation 1

F = 10(8)

F = 80 N.

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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
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a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

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Let's start by calculating the angular frequency:

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b)  t=0.10 s, t=0.52 s

The potential energy is given by:

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While the kinetic energy is given by:

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The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

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(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

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\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

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d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

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L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






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