16 when on increasing velocity
required
do work
what happens
to its
kinetic energy?
to

 Kinetic energy is proportional to the square of the velocity. If the velocity of an object doubles, the kinetic energy increases by a factor of four. Kinetic energy is proportional to the square

of the velocity. If the velocity of an object

doubles, the kinetic energy increases by a

factor of four.

• Kinetic energy is proportional to the mass. If

a bowling ball and a ping pong ball have the

same velocity, the bowling ball has much

larger kinetic energy.

• Kinetic energy is always positive.

• unit : Joule (J) = kg m

2

/s

2 Example:

If we drop a 3-kg ball from a height of h = 10 m,

the velocity when the ball hits the ground is

given by: v 2 = v0 2 +2a(y− y0 )= 0−2g(0−h)v= 2gh= 2(9.8 m/s 2 )(10 m)=14 m /s Initial:   k = 1 2 mv 2 = 0 Final:    k = 1 2 mv 2= 1 2 (3 kg)(14 m/s) 2= 294 J So as the ball falls, its kinetic energy increases. It is the gravitational force that accelerates the ball, causing the speed to increase. The increase in speed also increases the kinetic energy. The process of a force changing the kinetic energy of an object is called work. Work: Work is the energy transferred to or from an object by mean of a force acting on the object.• energy transferred to an object is positive work, e.g. gravity performs positive work on a

falling ball by transferring energy to the ball, causing the ball to speed up.• energy transferred from an object is negative work, e.g. gravity performs negative work on a ball tossed up by transferring energy from the ball, causing the ball to slow down.• both kinetic energy and work are scalars.• unit: J Work Energy Theorem: The work done is equal to the change in the kinetic energy: ∆K = K f − K i = W In the above example with the ball falling from a height of h = 10 m, the work done by gravity: W = ∆k = k f −k i = 294 J− 0J = 294 J. If a ball rises to a height of h =10 m, the work done by gravity: W = ∆k = k f −k i = 0J−294 J = −294 J. Work Done by a Force: Consider a box being dragged a distance d across a frictionless floor:

d F y x θ v 2 = v0 2 + 2ax (x − x0 ) v 2 = v0 2 +2ax d 1 2 mv 2 = 1 2 mv0 2 +max d 1 2 mv 2 − 1 2 mv0 2 = max d k f −k i = (Fcosθ)d ∴W = (Fcosθ)d• θ is the angle between the force vector and the direction of motion.• If the force is perpendicular to the direction of motion, then the work done: W =(Fcosθ)d = Fdcos90°= Fd×0= 0.• The work energy theorem and the relationship between work and force are valid only if the force does not cause any other form of energy to change, e. g. we can not apply the theorem when friction is

involved because it causes a change in the thermal energy (temperature). Work Done by Multiple Forces: The total work done by many forces acting on an object:Wtot = F1 cosθ 1 d+F2 cosθ 2 d+ F3 cosθ 3 d+L where the angles are the angle between each force and the direction of motion.  The total work is just the sum of individual work from each force:Wtot =W1 +W2 +W3 +L The work energy theorem relates the changes  in the kinetic energy to the total work performed on the object: ∆K =Wtot Example: A 3-kg box initially at rest slides 3 m down a frictionless 30° incline.  What is the work done on the object?  What is the kinetic energy and speed at the bottom?

x y N φ φ mg• The work done is performed by the force in the x direction since there is no motion in the y direction: W = F x d =(mgsinφ)d =(3 kg)(9.8 m/s 2 )(sin30°)(3 m) = 44 J Alternatively, W =(Fcosθ)d = Fcos(90°−φ)d = FsinφdH The first method of using the component of the force in the direction of motion for the calculation is easier.

there is no correct list among the choices you've provided.

the list might look something like this:

-- baby human

-- piccolo

-- flute

-- soprano singer

-- trumpet

-- tenor singer

-- trombone

-- baritone singer

-- baritone horn

-- bass singer

-- tuba

since i had to make up the list myself, i made it easy on myself, and i didn't try to shuffle the string instruments and woodwinds into it too.