Download Sam pushes a 10.0-kg sack of bread flour on a

PHYS 201
Exercise Set #16
Solutions
1
Problem 6.13 (5 points) Sam pushes a 10.0-kg sack of bread flour on a frictionless horizontal surface with a constant horizontal force of 2.0 N, starting from rest. (a) what is
the kinetic energy of the sack after Sam has pushed it a distance of 35 cm? (b) What
is the speed of the sack after Sam has pushed it a distance of 35 cm?
(a) We use the work-kinetic energy theorem. The only horizontal force is that of Sam’s
push, so the net force is just 2.0 N, in the direction of the displacement.
∆KE = Wnet
KE − KEo = F ∆x
KE = (2.0 N)(0.35 m)
=
0.70 J
(b) To find the speed, just use the definition of kinetic energy:
1 2
mv
2
1
0.70 J =
(10.0 kg)v 2
2
1.4 N-m
v2 =
10.0 kg
KE =
v =
0.37 m/s
Problem 6.15 (3 points) A ball of mass 0.10 kg moving with a speed of 2.0 m/s hits a wall
and bounces back with the same speed in the opposite direction. What is the change
in the ball’s kinetic energy?
We note that KE is 12 mv 2 . The speed v has the same magnitude but opposite sign.
This does not, however, change the quantity v 2 . There is no change in the ball’s KE.
Problem 6.18 (6 points) A plane weighting 220 kN (25 tons) lands on an aircraft carrier.
The plane is moving horizontally at 67 m/s (150 mi/h) when its tailhook grabs hold
of the arresting cables. The cables bring the plane to a stop in a distance of 84 m. (a)
How much work is done on the plane by the arresting cables? (b) What is the force
(assumed constant) exerted on the plane by the cables?
(a) By the work-kinetic energy principle, the work done on the plane must equal its
change in kinetic energy, Kf − Ki . Since the final kinetic energy is zero, the work done
is just −Ki . To find this, we need to know the mass, so we calculate it first:
m=
2.20×105 N
Fg
=
= 22,500 kg
g
9.8 m/s2
PHYS 201
Exercise Set #16
Solutions
2
Then
1
W = − mvo2
2
1
= − (22,500 kg)(67 m/s)2
2
= -5.0×107 J
(b) We use the definition of work, W = F x. Then
W
-5.04×107 J
=
x
84 m
5
= -6.0×10 N
F =
The negative sign just means the force is opposite the plane’s velocity.
Problem 6.19 (5 points) A shooting star is a meteor that burns up when it reaches Earth’s
atmosphere. Many of these meteors are quite small. Calculate the kinetic energy of
a meteor of mass 5.0 g moving at a speed of 48 km/s and compare it to the kinetic
energy of a 1100-kg car moving at 29 m/s (65 mi/h).
To find the kinetic energy of the meteor:
1 2
mv
2
1
=
(0.005 kg)(4.8×104 m/s)2
2
= 5.8×106 J or 5.8 MJ
KE =
For the car:
1 2
mv
2
1
=
(1100 kg)(29 m/s)2
2
= 4.6×105 J or 0.46 MJ
KE =
The kinetic energy of the tiny meteor (5 grams is the mass of a nickel) is 12.5 times
that of the more massive car.