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3 years ago

A boy im50kg at rest on a skateboard is pushed by another boy who exerts a force of 200 N on him. If the first boy's

Physics

1 answer:

sergey [27]3 years ago

5 0

Answer:

Time, t = 2 seconds

Explanation:

Given the following data;

Mass, m = 50 kg

Initial velocity, u = 0 m/s (since it's starting from rest).

Final velocity, v = 8 m/s

Force, F = 200 N

To find the time, we would use the following formula;

$F = \frac {m(v - u)}{t}$

Making time, t the subject of formula, we have;

$t = \frac {m(v - u)}{F}$

Substituting into the formula, we have;

$t = \frac {50(8 - 0)}{200}$

$t = \frac {50*(8)}{200}$

$t = \frac {400}{200}$

Time, t = 2 seconds

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3 years ago

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You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

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Answer:

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