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timofeeve [1]
3 years ago
11

When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a f

orce of 89 N, what is the magnitude of the horizontal component of the force that the floor exerts on the block?
Physics
1 answer:
elena-s [515]3 years ago
3 0

Answer:

Explanation:

When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .

It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply  a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

= 89 - 76

= 13 N

Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )

=

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0.28802

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Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

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The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

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Power to resistor

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Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

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