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3 years ago

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A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau

nched to a maximum height 50.2 cm. How much should the spring be compressed to send the ball twice as high?

1 answer:

AleksandrR [38]3 years ago

6 0

We know, by conservation of energy :

$\dfrac{kx^2}{2}=mgh$

Therefore,

$\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}$

Putting given values, we get :

$\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm$

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

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Answer:

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A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

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Answer:

$21.6\ \text{kg m}^2$

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$54.66\ \text{revolutions}$

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$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

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$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

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$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

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