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3 years ago

14

A 5.00 kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after an impulse with a magnitude o

f 30.0 newton • seconds is applied.

2 answers:

faltersainse [42]3 years ago

5 0

m = mass = 5 kg

$v_{i}$ = initial velocity = 100 m/s

$v_{f}$ = final velocity = ?

I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m( $v_{f}$ - $v_{i}$ )

30 = 5 ( $v_{f}$ - 100)

$v_{f}$ = 106 m/s

Romashka [77]3 years ago

5 0

The answer to this question is <span>106 m/s</span>

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3 years ago

A 75kg Tibetan is trekking along flat, but icy ledge with his 450kg yak when he slips over the edge. Luckily, he is holding the

Tcecarenko [31]

Answer:

minimal coefficient of static friction: $\mu_s=0.1667$

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* the normal force (indicated in green in the image and identified by the letter "n") of the ledge on the yak as reaction to the yak's weight

* the force of static friction between the yak's hooves and the ledge (pictured in red in the image and identified with $f_s$ )

Since the normal force and the force of gravity on the yak cancel each other (balance - the yak is not moving vertically), the only forces we need to analyse are the force of the Tibetan's weight via the strap, and the force of static friction which should at least be equal in magnitude so the Tibetan doesn't fall. We assume these two forces are acting horizontally (one to the right: the Tibetan's weight, and one to the left: the static friction).

As we said, we want them to be at least equal so thy are in balance.

We recall that the force of static friction is the product of the normal force (n) times the coefficient of static friction ( $\mu_s$ ), such that: $f_s=\mu_s\,*\,n$

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$F_g= M\,*\,g=450\, kg \,*\,g\\n=F_g=450\,kg\,*\,g\\f_s=\mu_s\,*\,n=\mu_s\,*450\,kg\,*\,g\\w=m\,*\,g=75\,kg\,*\,g$

So we need to find what is the minimum coefficient of static friction that precludes the Tibetan from falling. We therefore proceed to make an equality between the force of static friction on the yak and the weight of the Tibetan:

$f_s=w\\\mu_s\,*450\,kg\,*g=75\,kg\,*\,g$

and proceed to solve for the coefficient of friction by dividing both sides by "g" (which by the way cancels out), and by the yak's mass:

$\mu_s\,*450\,kg\,*g=75\,kg\,*\,g\\\mu_s=\frac{75}{450} \\\mu_s=0.1667$

where we have rounded to four decimal places the periodic number that the quotient generates. Notice that as expected, the coefficient of friction has no units (they all cancelled out in the division).

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