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2 years ago

You hold your physics textbook in your hand. (Assume that no other objects are in contact with the book.)

Physics

1 answer:

Jet001 [13]2 years ago

8 0

Answer:

1) the correct answer is b and d

2) For force b its direction is vertical up

for the force d its direction is vertical down

3) the correct answers are: a, c and d

4) Force a is vertical down , force c is vertical up and force d is vertical down

5) the correct answer which is b

Explanation:

In this exercise it is asked to identify the forces, fundamentally on the free there are the forces of gravity and the support force of the hand, with these facts we answer the questions

1) the correct answer is b and d

the hand acts on the book with a contact force and the Earth acts on the book with the force of gravity.

2) For force b its direction is vertical up

for the force d its direction is vertical down

3) The forces on the hand are the weight of the book. The force of gravity due to the mass of the hand. As the hand is in balance, there must be a force applied by the arm to keep the hand in position; assuming the hand is in the air, if the hand is resting on the floor the force of the floor on the hand can perform this function

therefore the correct answers are: a, c and d

4) Force a is vertical down

force c is vertical up

force d is vertical down

5) The action and reaction forces are forces of equal magnitude, each applied to one of the bodies, we have

* the force of the hand on the free and its reaction the force of the book on the hand

* The force of the Earth on the book and the hand, giving the weight of each one and the relationship is the force of the book and the hand on the Earth

the correct answer which is b

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A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert

pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

$\theta=65.4^{\circ}$

The projection of the length of the pendulum along the vertical direction is

$L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm$

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

$\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m$

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

$\Delta U = mg \Delta y$

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

$\Delta y = 0.211 m$ is the vertical displacement of the pendulum

So, the work done by gravity is

$W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J$

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement

$\theta$ is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

$\theta=90^{\circ}, cos \theta = 0$