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4 years ago

A cheerleader lifts his 59.6 kg partner straight up off the ground a distance of 0.749 m before

Physics

1 answer:

Airida [17]4 years ago

5 0

m = mass of the partner which the cheerleader lifts = 59.6 kg

h = height to which the partner is lifted by the cheerleader = 0.749 m

g = acceleration due to gravity = 9.8 m/s²

work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.

W = work done by the cheerleader in lifting the partner

PE = potential energy gained

so W = PE

potential energy is given as

PE = mgh

hence

W = mgh

inserting the values in the above formula

W = 59.6 x 9.8 x 0.749

W = 437.5 J

this is the work done in lifting the partner once.

the cheerleader does this 30 times , hence the total work done is given as

W' = 30 W

W' = 30 x 437.5

W' = 13125 J

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4 years ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

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3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

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where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

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5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

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3 years ago

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