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3 years ago

A 75.0 kg stunt man jumps from a balcony and falls 24.0

Physics

1 answer:

Anika [276]3 years ago

5 0

Answer:

The force that acts on the man equals 15354.75 newtons.

Explanation:

After falling through a distance of 24.0 meters the speed of the stunt man upon hitting the mattress can be obtained using third equation of kinematics as

$v^{2}=u^{2}+2gh\\\\\therefore v=\sqrt{2gh}\\\\v=\sqrt{2\times 9.81\times 24}\\\\v=21.7m/s$ ( u=0 since the man falls from rest)

Now since the man is decelerated through a distance of 1.15 meters thus the acceleration produced can be obtained from third equation of kinematics as

$v^{2}=u^{2}+2as\\\\0=u^{2}=2as\\\\a=\frac{-u^{2}}{2s}\\\\a=-\frac{(21.7^{2}}{2\times 1.15}=-204.73m/s^{2}$

Now by newton's second law the force that produced deceleration of the calculated magnitude is obtained as

$F=mass\times acceleration\\\\F=75.0\times -204.73=-15354.75Newtons$

The negative sign indicates that the direction of force is opposite of motion.

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3 years ago

A motorcycle stunt driver zooms off the end of a cliff at a speed of 41.9 meters per second. If he lands after 1.62 seconds, wha

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2 years ago

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

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Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance $d$ with the acceleration $a$ and time $t$ :