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den301095 [7]
3 years ago
14

A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.442 and t

he coefficient of kinetic friction is 0.240. How much force is needed to begin moving the book?
How much force is needed to keep the book moving at constant speed once it begins moving?
Physics
1 answer:
photoshop1234 [79]3 years ago
8 0

Answer:

a) 8.80 N

b) 4.78 N

Explanation:

The weight of the book is

W = mg = 2.03\times9.81 = 19.9143 \text{ N}

By Newton's third law, the reaction from the table, R, is equal to its weight.

R = 19.9143 N

From the law of solid friction,

F = \mu R

F is the maximum frictional force and \mu is the coefficient of friction.

a) The force needed to begin moving the book is that needed to overcome static friction

F = 0.442\times19.9143 = 8.80\text{ N}

b) The force needed to move the book at constant speed once it begins moving is that needed to overcome kinetic friction

F = 0.240\times19.9143 = 4.78\text{ N}

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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and
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Answer:

a) t= 0.92 s

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Explanation:

A)

  • In order to find the total time that the feet are in the air, we must add two times:
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  • 2) time from when starts to fall from the maximum height until her feet hit the water (t₂)
  • In order to get t₁, we need to take into account that at her highest point, the vertical speed will be zero.
  • Taking for granted the value for the acceleration due to gravity,
  • g = -9.8 m/s2, we can apply the definition of acceleration, and   replacing by the givens, we can find t₁ as follows:

       t_{1} = \frac{v_{o}}{g} = \frac{3.0m/s}{9.8m/s2} = 0.3 s (1)

  • In order to find t₂, we need to find first the highest point above the board, which is indeed what is asked for in b).
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  • The equation can be written as follows:

        v_{f} ^{2} - v_{o} ^{2} = 2*g*\Delta h  (2)

  • Replacing by the givens, and solving for Δh, we get:

       \Delta h = \frac{v_{o}^{2}}{2*g} = \frac{(3.0m/s)^{2} }{2*9.8m/s2} = 0.46 m (3)

  • The total height over the water will be just the sum of the takeoff point (1.40 m over the water) and the value we found in (3):
  • H = 1.40 m + 0.46 m = 1.86 m
  • Now, we can use the equation that relates the vertical displacement with the time, remembering that v₀=0, as follows:

       H = \frac{1}{2}*g*t^{2}  (4)

  • Replacing by the givens and the value found for H in (4), and solving for t, we get the value of t₂, as follows:

       t_{2} = \sqrt{\frac{2*H}{g} } = \sqrt{\frac{2*1.86m}{9.8m/s2} } = 0.62 s (5)

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B)

  • As we have just found, the highest point above the board was at 0.46m above the takeoff point, so the highest point above the board is just 0.46 m.

C)

  • In order to find the velocity when her feet hit the water, we can use the same equation (2), taking into account that v₀=0, and Δh = H= -1.86m.
  • Solving (2) for vf, we get:

        v_{f} = - \sqrt{2*g*H} = -\sqrt{2*9.8m/s2*1.86m} = -6.04 m/s (7)

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Answer:

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The law of conservation of energy states that "energy is neither created nor destroyed within a system but transformed from one form to another".

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