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Dafna11 [192]
3 years ago
13

A 100-kg athlete is performing a vertical leap to test his leg strength. He leaves the ground at 3.20 m/s, zooms upward to some

maximum altitude, and then returns to the ground. As he lands, he bends his legs and crouches down, so that he moves 30 cm downward as he comes to a stop. What average force did the athlete’s legs exert on the ground as he landed?
Physics
1 answer:
MaRussiya [10]3 years ago
5 0

Answer:

2687.7 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

When he will reach the ground his speed will be 3.2 m/s but will travel 30 cm down

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.2^2}{2\times 0.3}\\\Rightarrow a=-17.067\ m/s^2

Force

F=m(g-a)\\\Rightarrow F=100\times (9.81-(-17.067))\\\Rightarrow F=2687.7\ N

The force on the legs will be 2687.7 N

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An archer puts a 0.30-kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m. Assuming th
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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
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Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

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The diagram illustrating this is shown on the first uploaded

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   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

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         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

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          Tsin \theta  =   \frac{mv^2}{r}

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       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

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      a_c  =  \frac{v^2}{r}

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     a_c  =  9.8  *  tan (5.58)

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       T  = T_x i  + T_y  j

substituting value  

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