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gtnhenbr [62]
3 years ago
14

An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t

rue about this situation?
The athlete isn’t doing any work because he doesn’t move the weight.
The athlete isn’t doing any work because he doesn’t hold the weight long enough.
The athlete is doing work because he prevents the weight from falling downward.
The athlete is doing work because 50 kilograms is a significant load to lift.
Physics
2 answers:
Kamila [148]3 years ago
7 0

Answer: The true statement form the given statements is "the athlete is not doing any work because he does not move the weight.".

Explanation:

The formula for work done is as follows,

W = Fscos\theta

Here, F is the force, s is the displacement and theta is the angle between force and displacement.

In the given problem, an athlete holds 50 kg on his head for 10 seconds. The angle between force and displacement is 90 degree.

In this case, the work done is zero.

Therefore, the athlete is not doing any work because he does not move the weight.

Nataly_w [17]3 years ago
4 0
<span>The athlete is sweating, straining, sweating, trembling and groaning. 
But if he passed high school physics, then he may remember that
'work' in physics means (force) x (distance).  If there's no distance,
then there's no work. 

He realizes that he isn’t doing any work because he's not moving the weight.</span>
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Answer:

0.03167 m

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The potential energy and the kinetic energy of the system is conserved

P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m

The spring constant of the net is 20130.76 N

From Hooke's Law

F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m

The net would strech 0.03167 m

If h = 35 m

From energy conservation

65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0

Solving the above equation we get

x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45

The compression of the net is 1.52 m

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