57.

(a) With P = 1.5 MW = 1.5

× 10

6

W (assumed constant) and t = 6.0 min = 360 s, the work-kinetic

energy theorem (along with Eq. 7-48) becomes

W = P t = ∆K =

1

2

m

v

2

f

− v

2

i



.

The mass of the locomotive is then

m =

2P t

v

2

f

− v

2

i

=

(2)(1.5

× 10

6

W)(360 s)

(25 m/s)

2

− (10 m/s)

2

= 2.1

× 10

6

kg .

(b) With t arbitrary, we use P t =

1
2

m

v

2

− v

2

i



to solve for the speed v = v(t) as a function of time

and obtain

v(t) =



v

2

i

+

2P t

m

=



(10)

2

+

(2)(1.5

× 10

6

)t

2.1

× 10

6

=

√

100 + 1.5t

in SI units (v in m/s and t in s).

(c) Using Eq. 7-48, the force F (t) as a function of time is

F (t) =

P

v(t)

=

1.5

× 10

6

√

100 + 1.5t

in SI units (F in N and t in s).

(d) The distance d the train moved is given by

d =



t

0

v(t

) dt

=



360

0



100 +

3

2

t



1
2

dt =

4

9



100 +

3

2

t



3
2







360

0

which yields 6.7

× 10

3

m.