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gogolik [260]
8 months ago
13

A 2.0 kg book is lying on a 0.78 m -high table. You pick it up and place it on a bookshelf 2.1 m above the floor.Part ADuring th

is process, how much work does gravity do on the book?Part BDuring this process, how much work does your hand do on the book?
Physics
1 answer:
zloy xaker [14]8 months ago
7 0

Given data:

* The mass of the book is 2 kg.

* The initial height of the book is 0.78 m.

* The final height of the book is 2.1 m.

Solution:

(A). The work done by the gravity on the book is,

W=mg(h_f-h_i)

where m is the mass, g is the acceleration due to gravity, h_i is the initial height and h_f is the final height,

The work is done in moving the object in upward direction where as the gravitational force is acting in the down ward direction. Thus, the value g (acceleration due to gravity) is taken as negative in this case.

Substituting the known values,

\begin{gathered} W=2\times(-9.8)\times(2.1-0.78) \\ W=-19.6\times1.32 \\ W=-25.9\text{ J} \end{gathered}

Thus, the work done by the gravitational force is -25.9 J.

(B). The work done by the hand on the moving the book is,

\begin{gathered} W_1=-W \\ =25.9\text{ J} \end{gathered}

Thus, the work done by the hand on the book is 25.9 J.

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A 50-kg box is being pushed a distance of 8.0 m across the floor by a force P with arrow whose magnitude is 159 N. The force P w
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Answer:

Wp = 1,272 J

Wf = -940.8 J

Wn = 0

Wg = 0

Explanation:

  • Applying the definition of work, as the product of the component of the  force applied, along the direction of the displacement, times the displacement, we find that due to the normal force is always perpendicular to the surface, it does no work, as it has not a component in the direction of the displacement, so Wn = 0.
  • As the weight goes directly downward, it has no component in the direction of the displacement either, so Wg = 0.
  • We can get the work done by the force applied P, simply as follows:

        W_{p} = F_{p} * d * cos \theta = 159 N * 8.0 m * cos 0 = 1,272 J (1)

  • Finally, we have the work done by the force of friction that always opposes to the displacement, so it has negative sign.
  • The frictional force , can be written as follows:

        F_{fr} = \mu k * F_{n} (2)

        where μk = coefficient of kinetic friction = 0.24

         Fn = Normal Force

  • In this case, since the surface is level and horizontal, and there is no acceleration of the box in the vertical direction, this means that the normal force (in magnitude) must be equal to the weight:
  • Fn = m*g = 50 kg * 9.8 m/s2 = 490 N
  • Replacing these values in (2), we get:

       F_{fr} = 0.24 * 50 kg* 9.8 m/s2 = 117.6 N

  • Applying the definition of work, we can get the work done by the frictional force, as follows:

        W_{ffr} = F_{fr} * d * cos \theta = F_{fr} * d * cos (180) = - F_{fr} * d = -117.6 N * 8.0 m \\ =  W_{ffr} = -940.8 N

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Explanation:

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In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

Answer:

Ratio of length will be \frac{L_2}{L_1}=1.108

Explanation:

We have given time period of the pendulum when length is L_1 is T_1=0.95sec

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We know that time period is given by

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So 0.95=2\pi \sqrt{\frac{L_1}{g}}----eqn 1

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Dividing eqn 2 by eqn 1

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