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3 years ago

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A student is helping her teacher move a 9.5 kg box of books. what net sideways force must she exert on the box to slide it acros

s the floor so that it accelerates at 1.0 m/^2?

1 answer:

Tanzania [10]3 years ago

7 0

Mass of the box which contains books (m) = 9.5 kg

Acceleration required (a) = 1 m/s^2

Let the force applied by the teacher and student be F.

The only force acting in the horizontal direction is F and it will produce an acceleration of 1 m/s^2

According to Newton's Seconds law:

F(net) = ma

F = ma

F = 9.5 × 1

F = 9.5 Newtons

Hence, the sideways force applied by teacher and the students is equal to 9.5 Newtons

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To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

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$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

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k = Spring constant

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Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

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(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

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Answer:

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Given that,

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Hence, The time is 106.7 minute.

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