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Artyom0805 [142]
3 years ago
6

a pole vaulter at the top of his vault is 6.15 m in the air . if he has a gravitational potential energy of 4o42 j , what is his

mass?
Physics
2 answers:
Ksenya-84 [330]3 years ago
4 0
<span>82.0 kg I am going to assume that there is a typo for the number of joules of energy. Doing a google search for this exact question showed this question multiple times with a value of 4942 joules which makes sense given how close the "o" key is to the "9" key. Because of this, I will assume that the correct value for the number of joules is 4942. With that in mind, here's the solution. The gravitational potential energy is expressed as the mass multiplied by the height, multiplied by the local gravitational acceleration. So: E = MHA Solving for M, the substituting the known values and calculating gives: E = MHA E/(HA) = M 4942/(6.15*9.8) = M 4942/60.27 = M 81.99767712 = M Rounding to 3 significant figures gives 82.0 kg</span>
Rufina [12.5K]3 years ago
3 0

Gravitational potential energy is given by 

 U=mgh

 where 

 m is the mass of the object 

 h is the height of the object above the ground (Earth's surface)

 g is the gravitational acceleration constant, 9.8ms2.

 For h=6.15m and U=4042J, we can rearrange to solve for m:

 m=U/gh

 m=(4042J)/((9.8m/s^2)*(6.15m))

 m=67.06 Kg

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t= time= 60 ms= 0.06 s

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now force F= ΔP/t

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1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

F=mg=(7.6 kg)(9.8 m/s^2)=74.5 N

And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

F=kx\\k=\frac{F}{x}=\frac{74.5 N}{0.29 m}=256.9 N/m

2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s

And the frequency of oscillation is given by:

\omega=2\pi f\\f=\frac{2 \pi}{\omega}=\frac{2\pi}{5.81 rad/s}=1.08 Hz

3) 2.19 m/s

The velocity at time t of the block is given by:

v=v_0 cos (\omega t)

where

v_0 = 4.4 m/s is the initial velocity of the block

\omega=5.81 rad/s is the angular frequency

t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

v(0.36 s)=(4.4 m/s) ( cos ((5.81 rad/s)(0.36 s)) = -2.18 m/s

And the negative sign means that the direction of the velocity is upward (because the initial velocity was downward)

4) 25.6 m/s^2

The maximum acceleration is given by:

a_0 = \omega^2 A

where A is the amplitude of the oscillation.

We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m

So, the maximum acceleration is

a_0 = \omega^2 A=(5.81 rad/s)^2 (0.76 m)=25.6 m/s^2

5) 95.1 N

The magnitude of the net force acting on the block is given by the difference between the weight and the restoring force of the spring:

F=mg-kx

First, we need to find the position x at t=0.36 s, which is given by

x(t)=A sin(\omega t)=(0.76 m)(sin ((5.81 rad/s)(0.36 s))=0.66 m

And so, the net force is

F=(7.6 kg)(9.8 m/s^2)-(256.9 N/m)(0.66 m)=-95.1 N

And the negative sign means the direction of the force is upward.

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