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4 years ago

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A 77.0 kg rider sitting on a 7.3 kg bike is riding along at 9.3 m/s in the positive direction. The rider drags a foot on the gro

und and slows down to 6.0 m/s still in the positive direction.
A) What is the change in momentum of the rider and bike?

B) What is the impulse delivered by the ground to the rider's foot?

C) What force is acting on the bike and rider if slowing down took 13.2 seconds?

D) And, how far did the bike and rider travel during these 13.2 seconds?

1 answer:

saw5 [17]4 years ago

5 0

Answer with Explanation:

We are given that

Mass of rider=m=77 kg

Mass of bike =m'=7.3 kg

Initial velocity,u=9.3 m/s

Final velocity,v=6 m/s

A.Change in velocity=v-u=6-9.3=-3.3 m/s

Total mass,M=m+m'=77+7.3=84.3 kg

Change in momentum= $M(v-u)=84.3(-3.3)=-278.19 kgm/s$

B.Impulse=Ft=Change in momentum=-278.19kg m/s

C.Time,t=13.2 s

$v=u+at$

Using the formula

$6=9.3+13.2a$

$13.2a=6-9.3=-3.3$

$a=-\frac{3.3}{13.2}=-0.25 ms^{-2}$

F= $ma$

$F=84.3(-0.25)=-21.075 N$

D. $S=ut+\frac{1}{2}at^2$

$S=9.3(13.2)+\frac{1}{2}(-0.25)(13.2)^2$

$S=100.98 m$

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Pd = mgh÷Δt

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m = 250.0kg

g = 9.81m/s²

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Plug in the values and solve for Pd

Pd = 250.0×9.81×1.25÷0.206

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Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

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