Instructor: Dr. Bruce Law, CW
327, Tel: 532-1618.
REVIEW QUESTIONS OF Home Work 7
R5. The work required is the same i.e. Work
= mgd = 50 x 10 x 2 = 1000J for the first sack and
Work = mgd = 25 x 10 x 4 = 1000J for the second sack.
R9. It`s potential
energy would double because GPE = mgh where h is the
height, hence, if h is twice is large then GPE is twice as large.
R10. The heavier car has twice as much
potential energy as the lighter car because GPE = mgh
where m is the mass of the car.
R11. GPE = mgh = 1
x 10 x 4 = 40J at an elevation of 4m.
GPE = 80J at an elevation of 8m.
R13. KE = 0.5mv2 = 0.5 x 1 x 2 x
2 = 2J
R14. KE = 0.5mv2 therefore if v
increases by a factor of 4 then v2 = 4 x 4 = 16. Namely KE is 16
R19. Just before the apple hits the ground
the energy is converted to kinetic energy. When the apple hits the ground all
of it`s energy is converted to heat and sound energy.
EXCERISES OF Home Work 7
E1. It is easier to stop the lighter truck
because it`s KE is less, recall that KE = 0.5m v2
where m is the mass.
E13. The KE is maximum at the bottom of it`s swing where it`s PE is zero.
The PE is maximum at the top of it`s swing where it`s KE is zero. When the KE has half of it`s maximum value then the PE will have half of it`s maximum value as well so that the total energy E(tot) = PE + KE remains constant – this is just
conservation of total energy.
E26. Both balls will have the same speed according to the conservation of
total energy. This is because both tracks have the same height difference
between the beginning and the end of the track,
therefore, the same amount of gravitational potential energy is converted into
E28. The golf
ball and the Ping-Pong ball both have the same kinetic energy (where KE =
mvv/2). As the Ping-Pong ball has the smaller mass m, it must have the higher
speed v in order for it to have the same KE as the golf ball. Similarly for
light and heavy molecules with the same KE -- the lighter molecule must be
moving at a greater speed.
conservation of energy, the KE is converted into some other form of energy, in
this case, heat and sound energy.
E39. The stone
should penetrate twice as far because it starts off with twice as much GPE,
therefore it will have twice as much KE just before it hits the mud. The mud
does work on the stone in order to stop the stone where Work = Fd and F is the friction force
acting on the stone. As KE is twice as large the work to stop the stone must be
twice as large, therefore assuming the the friction
force F stays the same the penetration depth d must be twice as large.
PROBLEMS OF Home Work 7
P1. Work = Fd = 10 x 5 = 50J for the first case and Work = Fd = 20 x 2 = 40J for the first case. Assume that all of
this work is changed into KE, the first case must
produce the large KE.
does work to bring the car to a halt. By conservation of energy therefore, Work
= Fd = KE = mvv/2. If the
speed v is 3 times greater then the KE is 3x3=9 times greater, hence, the car
will skid 9 times as far i.e. 9x15 = 135m.