You hold your physics textbook in your hand. (assume that no other objects are in contact with the book.) (a) identify the forces acting on the book. (select all that apply.) book on hand hand on book floor on book earth on book incorrect: your answer is incorrect. for each system, consider only those forces outside of the system that act on the system. (b) for each force you identified in part (a), indicate the direction. (if a force is not applicable, choose na.) book on hand hand on book floor on book earth on book (c)

a) Hand on book and Earth (gravity) on book. b) Hand on book has a direction upwards and Earth on book has a direction downwards. Book on hand is not applicable and floor on book is not applicable.

Explanation:

In order to answer this question we must first do a drawing of the situation and a free body diagram (see picture attached).

a)

as you may see in the drawing, the book is only in contact with the hand and Earth applies a force on every object near its surface due to gravity, so the book is only affected by the contact it has with the hand (the normal force) and Earths's gravity (the weight). So the answer are: hand on book and Earth on book.

Book on hand is not an answer because we are not analyzing the hand, we are analyzing the book so we don't care about what forces the book is applying, we only care about which forces are being applied on the book. And floor on book is not applicable as well because the book is not in contact with the floor.

b) As you may see on the free body diagram, the force applied to the book due to gravity goes downwards, because the weight will always go in the direction of the center of the earth. We need a force that has the same magnitude and opposite in direction (due to Newton's third law of motion) so the normal force is given by the hand itself, so the force applied by the hand on the book goes upwards (oposite to its weight). The other forces don't apply because they are not affecting the book.

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

it's a b and