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3 years ago

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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the i

nstant when 28 cm of string has unwound off the disk, what is the torque exerted about the center of the disk?

1 answer:

drek231 [11]3 years ago

8 0

Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = $\frac{28}{10}$ × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

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Answer:

a) The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.

b) The initial translational kinetic energy of the bullet before the collision is 625 joules.

c) The final speed of the bullet-block system after the collision is 7.143 meters per second.

d) The total energy of the bullet-block system after the collision is 89.289 joules.

e) 89.289 joules must be done to stop the bullet-block system.

f) The bullet-block system will travel 13.007 meters before stopping.

Explanation:

a) Since no external forces are applied on the system defined by the bullet and the block, then the net momentum is conserved and can be calculated by the initial momentum of the bullet:

$p = m\cdot v_{o}$ (1)

Where:

$p$ - Net momentum, in kilogram-meters per second.

$m$ - Mass of the bullet, in kilograms.

$v_{o}$ - Initial speed of the bullet, in meters per second.

If we know that $m = 0.5\,kg$ and $v_{o} = 50\,\frac{m}{s}$ , then the net momentum of the bullet-block system before the collision is:

$p = (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)$

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The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.

b) The total energy of the bullet before the collision is its initial translational kinetic energy ( $K$ ), in joules:

$K = \frac{1}{2}\cdot m \cdot v_{o}^{2}$ (2)

$K = \frac{1}{2}\cdot (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)^{2}$

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The initial translational kinetic energy of the bullet before the collision is 625 joules.

c) Both the bullet and the block experiments a complete inelastic collision, then the final speed of the bullet-block system is calculated solely by the Principle of Momentum Conservation:

$v_{f} = \frac{m\cdot v_{o}}{m+M}$ (3)

Where:

$v_{f}$ - Final speed, in meters per second.

$M$ - Mass of the block, in kilograms.

If we know that $m = 0.5\,kg$ , $v_{o} = 50\,\frac{m}{s}$ and $M = 3\,kg$ , then the final speed of the bullet-block system is:

$v_{f} = \left(\frac{0.5\,kg}{0.5\,kg + 3\,kg} \right)\cdot \left(50\,\frac{m}{s} \right)$

$v_{f} = 7.143\,\frac{m}{s}$

The final speed of the bullet-block system after the collision is 7.143 meters per second.

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The total energy of the bullet-block system after the collision is 89.289 joules.

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$g$ - Gravitational acceleration, in meters per square second.

$s$ - Travelled distance, in meters.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

If we know that $m = 0.5\,kg$ , $M = 3\,kg$ , $\mu_{k} = 0.2$ , $g = 9.807\,\frac{m}{s^{2}}$ and $W = 89.289\,J$ , then the travelled distance of the bullet-block system is:

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$s = \frac{89.289\,J}{0.2\cdot (0.5\,kg + 3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$s = 13.007\,m$

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The formula for the mechanical energy E is:

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where K is the kinetic energy and U is the potential energy.

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