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PY1052 Problem Set 5 – Autumn 2004
Solutions
1.5 m
N
F worker
θ
θ
0.91 m
mg
(1) A 45-kg block of ice slides down a frictionless ramp 1.5 m long and
0.91 m high. A worker pushes up against the ice along the ramp so that
the block slides down at constant speed. (a) What is the force applied by
the worker? How much work is done by (b) the worker’s force? (c) the
gravitational force on the block? (d) the normal force on the block from
the ramp? (e) If the speed with which the block slides down is 0.5 m/s,
what is the power expended by the worker?
(a) Since the block slides with constant speed, there is no acceleration, and therefore
no net force. This means that the force of the worker up the ramp must be equal to
the component of the weight down the ramp:
Fworker − mg sin θ = 0
Fworker = mg sin θ = (45 kg)(9.8 m/s2 ) 0.91
= 268 N
1.50
(b) Wworker = Fworker d cos(180) = (268 N)(1.5 m)(−1) = −402 J
(c) Wgrav = mgd cos(90 − θ) = mgd sin θ = mgh = (45 kg)(9.8 m/s2 )(0.91 m) = 401 J
(d) Wnorm = N d cos(90) = 0
The normal force does no work, since it is perpendicular to the displacement
(e) Pworker = Fworker v cos(180) = (268 N)(0.50 m/s)(−1) = −134 J/s = 134 W
(2) The maximum force that you can exert on an object with one of your
teeth is about 750 N. Suppose that as you gradually bite on a piece of very
chewy licorice toffee, the toffee resists its compression with a spring-like
force of spring constant 2.5 × 105 N/m. Find (a) the distance the toffee is
compressed by your tooth and (b) the work your tooth does on the toffee
during the compression.
(a) F = kx −→ x =
F
k
=
750 N
2.5×105 N/m
= 3.0 × 10−3 m = 3.0 mm
(b) The work done by the spring-like force of the toffee is
Wspring = 21 kx2i − 12 kx2f
Wspring = 21 (2.5 × 105 N/m)[02 − (3.0 × 10−3 )2 ] = −1.125 J
Since there is no change in kinetic energy of the toffee just before and just after the
compression, there is no net work done. This means that the work done by your
tooth plus the work done by the spring-like force must be zero; therefore, the work
your poor tooth does on the toffee is +1.125 J.
(3) A 250-g block is dropped onto a relaxed vertical spring that has a
spring constant of k = 2.5 N/cm. The block compresses the spring 12 cm
before momentarily stopping. While the spring is being compressed, what
work is done on the block by (a) the gravitational force and (b) the spring
force? (c) What is the speed of the block just before it hits the spring
(assume we can neglect friction; solve part (c) using the work–kinetic
(a) The work by the gravitational force is
Wgrav = mgd cos(0) = mgd = (0.250 kg)(9.8 m/s2 )(0.12 m) = 0.294 J
(b) The work done by the spring force is
Wspring = 21 kx2i − 12 kx2f = 21 (2.5 × 102 N/m)[02 − (0.12 m)2 ] = −1.800 J
(c) The work–energy theorem says that the change in the kinetic energy is equal to
the total work done:
∆KE = Wgrav + Wspring
0 − 12 mvi2 = 0.294 J + (−1.800 J) = −1.506 J
vi2 =
2(1.506 J)
0.250 kg
= 12.05 m2 /s2
vi = 3.47 m/s
(4) A 4.50 kg cat is riding in a box with a mass of 0.5 kg attached to
a spring with spring constant k = 600 N/m. The cat is merrily sliding
over a horizontal frictionless surface. It has a kinetic energy of 35 J as
it passes through the spring’s equilibrium position. What is the total
power expended by the spring on the cat (a) as the cat passes through the
equilibrium position? (b) when the spring is compressed 0.10 m and the
cat is moving away from the equilibrium position? (c) when the spring is
compressed 0.05 m and the cat is moving toward the equilibrium position?
(a) The power expended by a force is P = F~ · ~v = F v cos φ, where φ is the angle
between the force and velocity. The magnitude of the spring force is F sp = kx, which
is zero at the equilibrium postion; therefore, the power expended by the spring force
as the cat passes through equilibrium is zero.
(b) Before finding the power, we must first find the spring force acting on the cat and
the velocity when the spring is compressed 0.10 m.
F = kx = (600 N/m)(0.10 m) = 60 N
We can use either the work–energy theorem or conservation of energy to find the
velocity. Using the work–energy theorem, we have
∆KE = Wspring = 21 kx2i − 21 kx2f
1
mv 2 − 35 J = 21 k(0)2 − 12 k(0.10 m)2
2
1
mv 2 = 35 J − 21 (600 N/m)(0.10 m)2 = 32 J
2
v=
r
2(32 J)
5.00 kg
= 3.58 m/s
v
F
We see above that the spring force and velocity are in opposite directions. Finally,
we must take into account the fact that the spring force expends some of its power to
move the box and some to move the cat. We see that the fraction of the total spring
force that acts on the cat is given by
Fsp
mcat +mbox
cat
= mcatm+m
Fsp
box
Fsp = (mcat + mbox )a −→ a =
sp
Fcat = mcat a = mcat mcatF+m
box
Fcat =
4.5
F
4.5+0.5 sp
= 0.90Fsp
Therefore, we have
Pcat = Fcat v cos φ = (0.90 ∗ 60 N)(3.58 m/s) cos(180) = −193 J/s = −193 W.
(c) Now we must first find the spring force and the velocity when the spring is compressed 0.05 m.
F = kx = (600 N/m)(0.05 m) = 30 N
As in (b), the force acting on the cat will be Fcat = 0.90 ∗ 30 N = 27 N. We can use
either the work–energy theorem or conservation of energy to find the velocity. Using
the work–energy theorem, we have
∆KE = Wspring = 21 kx2i − 21 kx2f
1
mv 2 − 35 J = 21 k(0)2 − 12 k(0.05 m)2
2
1
mv 2 = 35 J − 21 (600 N/m)(0.05 m)2 = 34.25 J
2
v=
r
2(34.25 J)
5.00 kg
= 3.70 m/s
F
v
We see above that the spring force and velocity are in the same direction. Therefore,
Pcat = Fcat v cos φ = (0.90 ∗ 30 N)(3.70 m/s) cos(0) = 99.9 J/s = 99.9 W.
(5) What is the spring constant of a spring that stores 25 J of elastic
potential energy when it is compressed by 7.5 cm?
The potential energy stored is P E = 12 kx2 ; therefore,
25 J = 21 k(0.075 m)2
k=
2(25 J)
(0.075 m)2
= 8.89 × 103 N/m
(6) An 6.00 kg stone is at rest on a spring. The spring is compressed
10.0 cm by the stone. (a) What is the spring constant? (b) If the spring is
compressed an additional 30.0 cm and then released, what is the maximum
height reached by the stone?
(a) Since there is no acceleration of the stone, the sum of the forces acting on it must
be zero; these are the spring force acting upward and its weight acting downward.
Thus,
kx − mg = 0 −→ k =
mg
x
=
(6.00 kg)(9.8 m/s2 )
0.10 m
= 588 N/m
(b) Use conservation of mechanical energy. We have both initially (when the spring
has been additionally compressed) and finally (whe the stone reaches its hightest
point) v = 0, so the kinetic energy at these points is also zero. Let’s take the height
when the spring is completely uncompressed to be height = 0 when calculating the
gravitational potential energy; this means that the initial height will be h i = −0.40 m.
We then have
(KE + P E)i = (KE + P E)f
0 + mghi + 12 kx2i = 0 + mghf + 12 kx2f
mghf = 12 k(x2i − x2f ) + mghi
mghf = 12 k[(0.40 m)2 − 02 ] − mg(0.40 m)
hf =
k
(0.40
2mg
m)2 − 0.40 m =
784 N/m
(0.40
2(8.00 kg)(9.8 m/s2
m)2 − 0.40 m = 0.40 m
The stone’s maximum height is 0.40 m above the uncompressed position of the spring.
(7) A 60 kg skier leaves the end of a ski-jump ramp with a velocity of
24 m/s directed 25◦ above the horizontal. The skier returns to the ground
at a point that is 14 m below the end of the ramp with a speed of 22 m/s.
(a) With what speed would the skier have landed if there were no air drag?
(b) By how much is the mechanical energy of the skier reduced by the air
drag?
(a) The angle at which the skier leaves the ski-jump is actually not needed. You can
treat her like a projectile and find her final y velocity and x velocity, then combine
them to get the total velocity. But this is not necessary – you can just use conservation
of mechanical energy in the absence of external forces or friction (air drag, in this
case):
(KE + P E)i = (KE + P E)f
1
mvi2 + mghi = 21 mvf2 + mghf
2
1
mvf2 = 12 mvi2 − mg(hf − hi )
2
vf2 = vi2 − 2g(hf − hi ) = (24 m/s)2 − 2(9.8 m/s2 )(−14.0 m) = 850 m2 /s2
vf = 29.2 m/s
(b) The reduction in the mechanical energy due to air drag can be found from the
relation
∆KE + ∆P E = Wdrag
( 12 mvf2 − 12 mvi2 ) + (mghf − mghi ) = Wdrag
1
m(vf2 − vi2 ) + mg(hf − hi ) = Wdrag
2
kg)[(22 m/s)2 − (24 m/s)2 ] + (60 kg)(9.8 m/s2 )(−14.0 m) = Wdrag
Wdrag = −1.10 × 104 J
The mechanical energy is decreased by 1.10 × 104 J
1
(60
2
Note that this is also the difference between her expected final kinetic energy when
mechanical energy is conserved and her actual final kinetic energy:
1
2
m(vf2 − vf,exp
) = 12 (60 kg)[(22 m/s)2 − (29.2 m/s)2 ] = −1.10 × 104 J
2
(8) A girl whose weight is 267 N slides down a 6.1 m playground slide
that makes an angle of 20◦ with the horizontal. The coefficient of kinetic
friction between the slide and the girl is 0.10. (a) How much energy is
transferred to thermal energy during her slide? (b) If she starts at the
top with a speed of 0.457 m/s, what is her speed at the bottom?
(a) The energy transferred to thermal energy will be the negative of the work done
by friction as she slides down the slide:
Wf ric = Fk d cos φ = (µk N )d cos(180) = −µk N d
We must find the normal force in order to find the work done by friction. We see that
the normal force is given by mg cos θ, where θ = 20◦ is the angle of the slide:
N
F
k
θ
mg
θ
θ = 20 deg
In the y direction, N − mg cos θ = 0 (no acceleration perpendicular to the slide).
Therefore
∆Eth = −Wf ric = µk N d = µk mg cos θd
∆Eth = (0.10)(267 N) cos(20)(6.1 m) = 153 J
(b) In the absence of external forces, we have
∆KE + ∆P E = −∆Eth
∆KE = −∆Eth − ∆P E
1
m(vf2 − vi2 ) = −∆Eth − mg(hf − hi )
2
vf2 = vi2 − m2 ∆Eth − 2g(hf − hi )
vf2 = (0.457 m/s)2 −
2
(153
27.2 kg
J) − 2(9.8 m/s2 (−6.1 m sin(20))
vf2 = 29.8 m2 /s2 −→ vf = 5.46 m/s
Above, we obtained the girl’s mass by dividing her weight by the gravitational acceleration.
Another way to think of this problem is that the final total mechanical energy is
equal to the initial total mechanical energy minus any energy that was transferred
into thermal energy:
KEf + P Ef = KEi + P Ei − ∆Eth
If you rearrange the terms, you’ll see that this is precisely the same as writing
∆KE + ∆P E = −∆Eth
(1b) The only force acting on a 2.0-kg canister that is moving in a horizontal xy plane has a magnitude of 5.0 N. The canister initially has a velocity
of 4.0 m/s in the +x direction and then some time later has a velocity of
6.0 m/s in the +y direction. How much work is done on the canister by
the 5.0 N force during this time?
The key thing here is that, since there is only one force acting on the canister in the
horizontal direction, the change in the canister’s kinetic energy must be due to the
work done by that force:
W = ∆KE = 12 m(vf2 − vi2 )
W = 12 (2.0 kg)[(6.0 m/s)2 − (4.0 m/s)2 ] = 20 J
The work done on the canister is 20 J. It is not important what direction the canister
is moving in initially and finally, since the kinetic energy depends only on the square
of the velocity. We also did not need to know how strong the force was – we needed
only to know the change in the canister’s kinetic energy.
(2b) A chain is held on a frictionless table with one-fourth of its length
hanging over the edge. If the chain has length L and mass m, how much
work is required to pull the hanging part back onto the table?
The total work that must be done is the total change in the gravitational potential
energy of the chain. If we assign the height of the table to be h = 0, the final
gravitational potential energy will be zero. The initial gravitational potential energy
of the 3/4 of the chain that is already on the chain is also zero. Thus, we must find
the gravitational potential energy of the 1/4 of the chain that is hanging.
Divide the hanging chain into a series of length elements dx. The mass of each of
dh. The gravitational potential energy of each element will be
these will be dm = m
L
m
dU = dm gh = L dh gh = mg
h dh.
L
Now just integrate over all possible h values for the hanging part of the chain:
PE =
R0
mg
mg R 0
−L/4 L h dh = L −L/4
mg 1 2 0
1
h |−L/4 = − 32
mgL
L 2
h dh
PE =
This is the total initial gravitational potential energy of the chain; since the total
final gravitational potential energy is zero, the work that must be done is given by
1
1
∆P E = Wext −→ 0 − (− 32
mgL) = 32
mgL
We see that this is also equal to the amount of work needed to lift a “particle” with a
mass of m/4 located at the centre of mass of the hanging chain, at a height of −L/8
(below the table top), onto the table. The change in gravitational potential energy
in this case is
1
) = 32
mgL
∆P E = 0 − m4 g( −L
8
(3b) The cable of a 1800 kg elevator cabin snaps when the cabin is at rest
a distance d = 3.7 m above a cushioning spring whose spring constant is
k = 0.15 MN/m. A safety device clamps the cap against guide rails so that
a constant frictional force of 4.4 kN opposes the cab’s motion. (a) What
is the speed of the cabin just before it hits the spring? (b) What is the
maximum distance the spring is compressed? (c) What is the distance the
cabin will bounce back up the shaft? (d) Estimate the total distance that
the cab will move before coming to rest.
(a) Find the speed just before hitting the spring from the relation
∆KE + ∆P E = Wf ric
1
m(vf2 − vi2 ) + mg(hf − hi ) = Wf ric
2
vf2 = vi2 − 2g(hf − hi ) + m2 Wf ric = −2g(hf − hi ) − m2 Ff ric (hi − hf )
vf2 = −2(9.8 m/s2 )(−3.7 m) −
2
(4.4
1800 kg
× 103 N)(3.7 m)
vf2 = 72.5 − 18.1 = 54.4 m2 /s2
vf = 7.38 m/s
Note that hf −hi is negative because height increases upward; also Wf ric = Ff ric d cos(180) =
−Ff ric (hi − hf ).
(b) We can get the maximum distance the spring is compressed in the same way, but
including the change in both the gravitational and spring potential energy in ∆P E.
We’ll take the initial time to be when the cab first hits the spring and the final time
to be when the spring is maximally compressed (vf = 0):
∆KE + ∆P E = Wf ric
1
m(vf2 − vi2 ) + mg(hf − hi ) + 21 k(x2f − x2i ) = Wf ric
2
1
m(0 − vi2 ) + mg(−xf ) + 12 kx2f = −Ff ric xf
2
Here, x is the amount the spring is compressed (or extended); x i = 0, and hf −
hi = −xf (the height decreases by the amount the spring is compressed). Solve this
quadratic equation for xf :
1
kx2f + (Ff ric − mg)xf − 12 mvi2 = 0
2
( 21 1.50×105 N/m)x2f +(4.4×103 N−(1800 kg)(9.8 m/s2 ))xf − 12 (1800 kg)(7.38 m/s)2 = 0
(7.50 × 104 N/m)x2f − (1.32 × 104 N)xf − 4.90 × 104 J = 0
√
1.32×104 N± (−1.32×104 N)2 −4(7.50×104 N/m)(−4.90×104 J)
xf =
2(7.50×104 N/m
4
5
N±1.22×10 N
xf = 1.32×10
= 0.90 m, −0.725 m
1.50×105 N/m
The value we want is xf = +0.90 m, since we set up our work–energy equation above
assuming xf was positive.
(c) Find the distance the cabin bounces back up the shaft in precisely the same way;
note that both vi and vf are zero:
∆KE + ∆P E = Wf ric
1
m(vf2 − vi2 ) + mg(hf − hi ) + 21 k(x2f − x2i ) = Wf ric
2
mg(hf − hi ) + 21 k(0 − x2i ) = Wf ric
mghf = mghi + 12 kx2i + Wf ric
mghf = mghi + 12 kx2i − Ff ric (−xi + hf )
(mg + Ff ric )hf = mghi + 21 kx2i − Ff ric (−xi )
[(1800 kg)(9.8 m/s2 ) + 4.4 × 103 N]hf =
(1800 kg)(9.8 m/s2 )(−0.90 m) + 12 (1.5 × 105 N/m)(0.90 m)2 −
(4.4 × 103 )(0.90 m)
(2.20 × 104 )hf = −1.59 × 104 + 6.08 × 104 − 3.96 × 103 = 4.09 × 104
hf = 1.86 m
The lift bounces back up to a height of 1.86 m above the uncompressed position of the
spring cushion. We can see that this height is lower than the initial height, as it should
be, since some of the initial gravitational potential energy has been transformed into
thermal energy by friction.
(d) Since the friction force is constant and always acts opposite to the direction of
motion, the total work done on the system by the friction force will just be W tot =
−Ff ric dtot , where dtot is the total distance the cabin moves before finally coming to
rest. This will also be equal to the total change in the mechanical energy of the
system:
∆KE + ∆P E = Wf ric
∆P E = Wf ric
since the initial and final kinetic energy are zero
To find the total change in gravitational potential energy, we must simply find the
final height of the cabin. This is given by
kxf − mg = 0 −→ xf = mg
k
2
kg)(9.8 m/s )
= 0.12 m
xf = (1800
1.5×105 N/m
Taking h = 0 to be the uncompressed position of the spring, the initial height is
hi = 3.70 m and the final height is hf = −0.12 m. Therefore,
∆P E = Wf ric
mg(hf − hi ) = −Ff ric dtot
dtot =
mg(hi −hf )
Ff ric
=
(1800 kg)(9.8 m/s2 )(3.70 m+0.12 m)
4.4×103 N
= 15.3 m