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3 years ago

What average force is necessary to bring a 50.2-kg sled from rest to a speed of 2.8 m/s in a period of 20.1 s?

1 answer:

7 0

To solve this problem we will use the concepts related to the Impulse-Momentum Theorem for which it is specified as the product between force and change in time

$\Delta p = F\Delta t$

And

\Delta p = m\Delta v

Where,

$F = Force$

$\Delta t = \text{Change in Time}$

$\Delta v = \text{Change in velocity}$

$m = mass$

Rearranging to find the Force we have that

$F = \frac{\Delta p}{\Delta t}$

Using the expression between mass and velocity

$F = \frac{m(v_f-v_i)}{\Delta t}$

Our values are given as,

$m = 50.2kg\\v_i = 0m/s \\v_f = 2.8m/s \\\Delta t = 20.1s$

Then replacing we have that

$F = 6.99N$

Therefore the average force is 6.99N

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b) See plot attached

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d) 0.500 cm

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The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

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t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

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y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

$\lambda=2L$