Preparation problems

7-48. At what speed must a 950-kg subcompact car be moving to have the same kinetic energy as a 3.2 ×104-kg truck going 20 km/h?

Solution:

We start by converting the speed of the truck into MKS units:

vT = 20  km/h = 20  km/h 1000  m/km
3600  s/h
 = 5.56 m/s

Now we can find the kinetic energy of the truck, KT:

KT
=
1
2
 mT vT2
=
1
2
 (3.2 ×104  kg) (5.56  m/s)2
KT
=
493827  J

Now we can use the definition of kinetic energy to solve for the speed of the car, vc, at which it has this much kinetic energy:

KT
=
1
2
 mc vc2
vc2
=
2 KT
mc
vc
=
  æ
Ö
2 KT
mc
 
=
  æ
Ö
2 (493827  J)
(950  kg)
 
vc
=
32.2  m/s

7-52. After a tornado, a 0.50-g drinking straw was found embedded 4.5 cm in a tree. Subsequent measurements showed that the tree would exert a stopping force of 70 N on the straw. What was the straw's speed when it hit the tree?

Solution:

We can solve this problem using the Work-Energy theorem:

Wnet = DK
While in the tree, the net force acting on the straw was the 70-N stopping force. The work done by this stopping force is therefore the net work. This is a constant force acting in one dimension, so we do not have to resort to the integral definition of work, and can use the simple form
Wnet
=
Fnet Dx
=
(-70  N) (0.045  m)
Wnet
=
- 3.15  J
This net work must equal the change in the kinetic energy of the straw, or
Wnet = Kf - Ki
Since the straw stopped, Kf = 0 J, and we can solve for the initial velocity of the straw using the definition of kinetic energy.
Wnet
=
- Ki
Wnet
=
- 1
2
 mvi2
vi2
=
- 2Wnet
m
vi
=
  æ
Ö
- 2Wnet
m
 
=
  æ
Ö
- 2 (- 3.15  J)
0.5 ×10-3  kg
 
vi
=
112.25  m/s

7-67. In midday sunshine, solar energy strikes Earth at a rate of about 1 kW/m2. How long would it take a perfectly efficient solar collector of 15 m2 area to collect 40 kWh of energy?

Solution:

If solar energy strikes the earth at a rate of 1 kW/m2, a 15-m2 collector should gather energy at a rate of

P = (1  kW/m2)(15  m2) = 15 kW
Since we know by definition
P = dW
dt
 = DE
Dt
 ,
we see then that
DE
=
P Dt
Dt
=
DE
P
=
40 kWh
15 kW
Dt
=
2.67 h

7-70. Which consumes more energy, a 1.2-kW hair dryer used for 10 min or a 7-W night light left on for 24 h?

Solution:

Since we know by definition

P = dW
dt
 = DE
Dt
 ,
we see then that
DE = P Dt
So we can solve for the energy used by each appliance. First the hair dryer:
DE
=
P Dt
=
(1200  W) (10  min 60  s
1  min
 )
DE
=
720 kJ

Now the night light:

DE
=
P Dt
=
(7  W) (24  h 3600  s
1  h
 )
DE
=
605 kJ

So we can see that the hair dryer used more energy.

Follow-up problems

7-30. A spring has a spring constant k = 200 N/m. How much work does it take to stretch the spring (a) 10 cm from equilibrium and (b) from 10 cm to 20 cm from equilibrium?

Solution:

We start out again by using our definition of work:

W = ó
õ 		r2

r1 
 ®
F
 
 ·d ®
r
 
We are operating in one dimension, so we can drop the vector symbols. Since the force with which we are dealing is not constant, we must perform the integration to solve the problem:
W = ó
õ 		x2

x1 
 F(x) dx

(a) To find the work done, we must calculate the integral

W = ó
õ 		0.1  m

0  m 
 kx dx
Since k is constant, we can pull it outside the integral and continue
W
=
k ó
õ 		0.1  m

0  m 
 x dx
=
k   1
2
 x2 ê
ê
ê 		0.1  m

0  m 
=
1
2
 (200  N/m) [ (0.1  m)2 - (0  m)2 ]
W
=
1 J

(b) To find the work done, we must calculate the integral

W = ó
õ 		0.2  m

0.1  m 
 kx dx
Since k is constant, we can pull it outside the integral and finish
W
=
k ó
õ 		0.2  m

0.1  m 
 x dx
=
k   1
2
 x2 ê
ê
ê 		0.2  m

0.1  m 
=
1
2
 (200  N/m) [ (0.2  m)2 - (0.1  m)2 ]
W
=
3 J

7-35. A force given by F = a Öx acts in the x-direction, where a = 14 N/m1/2. Calculate the work done by this force acting on an object as it moves (a) from x = 0 to x = 2.2 m; (b) from x = 2.2 m to x = 4.4 m; (c) from 4.4 m to 6.6 m.

Solution:

We start out again by using our definition of work:

W = ó
õ 		r2

r1 
 ®
F
 
 ·d ®
r
 
We are working only in the x-direction, so we can drop the vector symbols. Since the force with which we are dealing is not constant, we must perform the integration to solve the problem:
W = ó
õ 		x2

x1 
 F(x) dx

(a) To find the work done, we must calculate the integral

W = ó
õ 		2.2  m

0  m 
 aÖx dx
Since a is constant, we can pull it outside the integral and continue
W
=
a ó
õ 		2.2  m

0  m 
 Öx dx
=
a   2
3
 x3/2 ê
ê
ê 		2.2  m

0  m 
=
2
3
 (14  N/m1/2) [ (2.2  m)3/2 - (0  m)3/2 ]
W
=
30.45  J

(b) Again, to find the work done, we must calculate the integral

W = ó
õ 		4.4  m

2.2  m 
 aÖx dx
Since a is constant, we can pull it outside the integral and finish
W
=
a ó
õ 		4.4  m

2.2  m 
 Öx dx
=
a   2
3
 x3/2 ê
ê
ê 		4.4  m

2.2  m 
=
2
3
 (14  N/m1/2) [ (4.4  m)3/2 - (2.2  m)3/2 ]
W
=
55.7  J

(c) Finally, to find the work done, we must calculate the integral

W = ó
õ 		6.6  m

4.4  m 
 aÖx dx
Again, a is constant and we can pull it outside the integral and finish
W
=
a ó
õ 		6.6  m

4.4  m 
 Öx dx
=
a   2
3
 x3/2 ê
ê
ê 		6.6  m

4.4  m 
=
2
3
 (14  N/m1/2) [ (6.6  m)3/2 - (4.4  m)3/2 ]
W
=
72.1  J

Physics 110 Home Page Physics 110 Honors Home Page

Solutions translated from TEX by TTH, version 1.57.