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2 years ago

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A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in

motion, in horizontal force of 650N

1 answer:

miv72 [106K]2 years ago

3 0

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is $\mu_s = 0.60$

The value for static friction is $\mu_k = 0.70$

Explanation:

From the question we are told that

The mass of the clock is $m = 95 \ kg$

The first horizontal force is $F _s = 560 \ N$

The second horizontal force is $F _k = 650 \ N$

Generally the static frictional force is equal to the first horizontal force

So

$F _s = m * g * \mu_s$

=> $560 = 95 * 9.8 * \mu_s$

=> $\mu_s = 0.60$

Generally the kinetic frictional force is equal to the second horizontal force

So

$F _k = m * g * \mu_k$

$650 = 95 * 9.8 * \mu_k$

$\mu_k = 0.70$

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(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

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$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

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$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

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$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

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(e) 155 kg

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$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=\frac{B}{g}-m=\frac{4159 N}{9.8 m/s^2}-269 kg=155 kg$

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As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

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So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

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