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Lynna [10]
3 years ago
10

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

the coefficient of kinetic friction?
Physics
2 answers:
andrey2020 [161]3 years ago
5 0
Considering that the book is moving with constant speed, the force applied by Anna must be the same that the friction force:

F = F_R = F\cdot \mu_k\cdot N

If we clear the previous equation:

\mu_k = \frac{F}{F_R} = \frac{19.5\ N}{51.7\ N} = \bf 0.38
barxatty [35]3 years ago
5 0
The force applied to the book is equal to the normal force of the book multiplied by the coefficient of kinetic friction. <span>The normal force is equal to the weight of the book. </span>In this case, the formula with the given data is equal to 19.5N = 51.7N * coeff. of friction. The coefficient of friction is equal to 0.38.
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pantera1 [17]

Answer:

The x-component of F_{3} is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O (1)

Where:

\vec F_{1}, \vec F_{2}, \vec F_{3} - External forces exerted on the ring, measured in newtons.

\vec O - Vector zero, measured in newtons.

If we know that \vec F_{1} = (70.711,70.711)\,[N], \vec F_{2} = (-126.859, 46.173)\,[N], F_{3} = (F_{3,x},F_{3,y}) and \vec O = (0,0)\,[N], then we construct the following system of linear equations:

\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N (2)

\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N (3)

The solution of this system is:

F_{3,x} = 56.148\,N, F_{3,y} = -116.884\,N

The x-component of F_{3} is 56.148 newtons.

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3 years ago
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sasho [114]

Let's choose the "east" direction as positive x-direction. The new velocity of the jet is the vector sum of two velocities: the initial velocity of the jet, which is

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