Download wor, power, energy

Workshop Tutorials for Biological and Environmental Physics
Solutions to MR5B: Work, Power and Energy
A. Qualitative Questions:
1. The total work done on the TV really is zero. It starts with no kinetic energy and ends with no kinetic
energy. However this does not mean that Brent does no work on the TV. He must accelerate it to move it,
and increase its potential energy, doing work against gravity, if he lifts it. This energy comes from
chemical potential energy stored in Brent. When he decelerates the TV and puts it down he must absorb
the energy he has put in. However this energy is not converted back to chemical potential energy, but is
lost as heat. In addition, if Brent slides the TV then the energy he has put in to accelerate it and give it
kinetic energy will be dissipated as heat due to friction by the floor and TV. So while the energy of the
TV has not changed, and no work has been done on it, Brent must do work to move it.
2. No net work is done on a body in uniform circular motion. The net force (or centripetal force) is
always directed in towards the centre of the circle, and is at right angles to the direction of travel. Hence
the work W = F.d = Fdcos = 0, because the force, F, is perpendicular to the direction of the
displacement, d. Using energy considerations we come to the same conclusion – neither potential energy
nor the kinetic energy of the body is changing, hence no work can be being done on the body.
The Earth does not follow a circular path, however over a complete orbit there is still no net work done.
The Earth travels in an elliptical path, so the gravitational force between the sun and Earth is not quite
perpendicular to the Earth’s path. The Earth speeds up as it moves closer to the sun, and gravity does
work. It slows down again as it moves further away. There is a variation in the gravitational potential
energy and kinetic energy of the Earth, however the total energy is always constant. Over a complete orbit
the gravitational force does no net work.
B. Activity Questions:
1. Pendulum
The forces acting on the pendulum are its weight (gravity), and the tension in
the string. The tension is always at right angles to the path, hence it does no
work. Ignoring friction, only the weight of the pendulum does work as it
swings, converting gravitational potential energy into kinetic energy and back
again.
T
mg
2. Falling
When you drop an object it falls due to gravity, losing gravitational potential energy. As it falls this
gravitational potential energy is converted to kinetic energy, and the object gains speed. The change in
kinetic energy is equal to the work being done on the falling object, and is done by gravity. Air resistance
may also do some negative work on the object, acting to reduce its kinetic energy and slow it down.
3. Power
The power used by the appliances is written on the back, and is measured in watts, W, or sometimes volts
 current or VI. If an appliance is rated at X watts it converts X joules per second of energy.
A hairdryer converts electrical energy into thermal energy (heat) and kinetic energy, a lamp produces heat
and light. All appliances convert at least some electrical energy into thermal energy.
The Workshop Tutorial Project –Solutions to MR5B: Work, Power and Energy 87
C. Quantitative Questions:
1. A worker ant is carrying a grain of sand weighing 1.0 g out of the
nest to add to the pile. When he reaches the nest entrance the ant drops
the sand and pushes it, at constant velocity, up the slope surrounding the
nest entrance. This slope makes an angle of 40o to the horizontal and is
5 cm long. The coefficient of friction between the slope and the sand is
Fpush
0.97.
a. If the ant pushes at constant velocity, then the force, Fpush, he exerts
up the slope equals the net force down the slope. The forces down the
slope are the component of the gravitational force, mgsin and the
frictional force Ffriction =N =mgcos. Therefore
Ffriction
Fpush = mgsinmgcos = mg(sin cos)
Fgravity
40o
= 1.0  10-9 kg  9.8 m.s-2 (sin 40o + 0.97  cos40 o) = 13.58 x 10-9 N.
b. Work of the ant on the sand is Want = Fpushdcos  =13.58  10-9 N  0.05 m cos(0 o) = 6.79 x10-8 J.
c. Work done by the weight of the sand,
Wgravity = Fgravity d cos = mgdcos = 1.0  10-9 kg  9.8 m.s-2  0.05 m  cos(130o) = -3.15  10-8 J.
d. Work done by friction,
Wfriction = Ffriction dcos = Ndcos =mgcos dcos(180o)
= 0.97  (1.0  10-9 kg  9.8 m.s-2  cos(40o) ) 0.05 m  cos (180o) = -3.64  10-8 J.
e. The total work is sum of b, c and d above.
Total work = Want + Wgravity +Wfriction = 6.79 x10-8 J + -3.15  10-8 J + -3.64  10-8 J = 0 J.
We expect this zero value as the kinetic energy is unchanged (Work-Energy theorem).
2. To stay hovering a bee beats its wings up and down, with the wings angled differently on the up and
down stroke.
a. On the down stroke the air resistance (drag) will be large as the maximum cross section moves
through the air. As the insect pushes down (action) the air pushes back (reaction). If the push up by the air
on the insect is greater than the gravitational force the acceleration will be upwards. On the upstroke the
wing is angled so the air resistance is much much less and therefore the acceleration will be
approximately gravitational. The bee is in free fall during the upstroke and gains kinetic energy as it falls.
b. During the down stroke the bee gains gravitational potential energy. If we ignore any frictional losses
then work done = KE lost = GPE gained.
Work done = mgh = 0.1 x10-3 kg  9.8 m.s-2  1.0  10-3 m = 9.8 x10-7 J
c. We assume that the work done by the bee’s wings on the upstroke is zero as the wings provide no air
resistance and the bee is in free fall.
d. The power generated is the rate of work done. In
The power is 9.8 x10-7 J /
1
200
1
200
s, 9.8  10-7 J of work is done.
= 9.8 x10-7 J  200 s-1=1.96  10-4 J.s-1 = 0.20 mW.
88 The Workshop Tutorial Project –Solutions to MR5B: Work, Power and Energy