13–2.
v = 10 ft/s

The 10-lb block has an initial velocity of 10 ft>s on the smooth
plane. If a force F = 12.5t2 lb, where t is in seconds, acts on the
block for 3 s, determine the final velocity of the block and the
distance the block travels during this time.

F = (2.5t) lb

SOLUTION
10

+

2.5t = ¢ 32.2 ≤a

: ©Fx = max;

a = 8.05t
dv = a dt
n

L10

t

dv =
L0 8.05t dt
2


United
use

When t = 3 s,
for

instructors World permitted

of learning the is not
on
and

the student
(including work

by

s = 66.2 ft
protected
is

of the

assessing
solely

work

work provided and of



SOLUTION
Free-Body Diagram: The kinetic friction Ff = mkN is directed to the left to oppose the
motion of the crate which is to the right, Fig. a.
Equations of Motion: Here, ay = 0. Thus,
+ c ©Fy = 0;

N - 50(9.81) + 200 sin 30° = 0
N = 390.5 N

+
: ©Fx = max;

200 cos 30° - 0.3(390.5) = 50a
2

a = 1.121 m>s

Kinematics: Since the acceleration a of the crate is constant,
+

laws

:B

A

v = v0 + act

teaching

.

copyrightAns. Wide
Dissemination
.
States
World permitted

and
+

or

on

by

2

0 + 0 + 2 (1.121) A3 B = 5.04 m

the student work
(including
protected
of the
is
solely
work
assessing
this


If the 50-kg crate starts from rest and achieves a velocity of v =
4 m>s when it travels a distance of 5 m to the right, determine
the magnitude of force P acting on the crate. The coefficient of
kinetic friction between the crate and the ground is mk = 0.3.

308

SOLUTION
Kinematics: The acceleration a of the crate will be determined first since its motion is
known.
:

+

( )

2 =

v

2 +

-

v0
2

2ac(s


.

+
: ©Fx

States
= max;

P cos 30° - 0.3(490.5 - 0.5P)

United

P = 224 N

World permitted
not
of learning the is
on andAns.
instructors

= 50(1.60)
use

by
the student
(including work
protected
of the
is
solely

fast the sled is traveling when s = 5 ft.
100 ft

SOLUTION
+ b aFx = max;

800
a
32.2

800 sin 45° - 30 =

100 ft

s

2

a = 21.561 ft>s
2

2

v1 = v0 + 2ac(s - s0)
v1

2

= 0 + 2(21.561)(100 2


Dissemination or

copyright

+ 2( -3.22)(5 - 0)

+ 2ac(s2 - s1)

permitted

.

instructors
States

v2

Wide

of

United

= 77.9 ft>s

World
learning
on the

use


and courses part
any
of
their

des troy

sale will

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,


13–6.
If P = 400 N and the coefficient of kinetic friction between the
50-kg crate and the inclined plane is mk = 0.25, determine the
velocity of the crate after it travels 6 m up the plane. The crate
starts from rest.

P

30°

SOLUTION

30°

Free-Body Diagram: Here, the kinetic friction Ff = mkN = 0.25N is required to be

States

2

United

2

.

instructors World permitted

of learning the is

Kinematics: Since the acceleration a of the crate is constant,

use

v = v0 + 2ac(s - s0)

and
student

2

by the
v

not


any
of
their

des troy

sale will

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by
Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system,
or transmission in any form or by any means, electronic, mechanical,


13–7.
If the 50-kg crate starts from rest and travels a distance of 6 m
up the plane in 4 s, determine the magnitude of force P acting
on the crate. The coefficient of kinetic friction between the
crate and the ground is mk = 0.25.

P

30°

SOLUTION

30°

Kinematics: Here, the acceleration a of the crate will be determined first since its
motion is known.
s = s0 + v0t +


Wide

plane, Fig. a.

©Fy¿ = may¿;

.

N + P sin 30° - 50(9.81) cos 30° =

50(0)

N = 424.79 - 0.5P

instructors
permitted
States
. World
learning on
of

United

and

use

Using the results of N and a,
©Fx¿ = max¿;

their

des troy

sale will

© 2013 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by


*13–8.
The speed of the 3500-lb sports car is plotted over the 30-s time
period. Plot the variation of the traction force F needed to cause
the motion.

v(ft/s)
F

80
60

SOLUTION
Kinematics: For 0 … t
we have

6 10 s. v =

60
dv
10 t = {6t} ft>s. Applying equation a = dt ,



or

teaching Web)
Dissemination
copyright
Wide
.

Equation of Motion:
For 0 … t 6 10 s
;

+

aFx = max ; F =

¢

3500
32.2 ≤(6) =

instructors
permitted
States
. World
learning

of
United

assessing

109 lb

is

solely

work provided and of

This is

work

Ans.

integrity
this

the

and courses part
any
of
their

des troy

sale will


P cos 20° - 0.5N = 0
Solving Eqs.(1) and (2) yields
P = 353.29 N
Equations of Motion: The friction

(2)

N = 663.97 N
force developed between

the crate and its
laws

or

contacting surface is Ff = mkN = 0.3N since the crate is moving. From FBD(b),

+ c ©Fy

= may ;

: ©Fx = max ;
+

teaching Web)
Dissemination
copyright
Wide
.



Aisns
.

and

the student
for
(including work
by

protected

of the
assessing

is

solely

work provided and of

This is

work

integrity
this

the


Equations of Equilibrium: At t = 2 s, P = 90 A 2 B = 360 N. From FBD(a)
+ c ©Fy = 0;

N + 360 sin 20° - 80(9.81) = 0

N = 661.67 N

+
360 cos 20° - Ff = 0Ff = 338.29 N
: ©Fx = 0;
Since Ff 7 (Ff)max = ms N = 0.4(661.67) = 264.67 N, the crate accelerates.
Equations of Motion: The friction force developed between the crate and its contacting
surface is Ff = mkN = 0.3N since the crate is moving. From FBD(b),
+ c ©Fy = may ; :

+ N - 80(9.81) + 360 sin 20° = 80(0)
N = 661.67 N

laws

Web)

teaching

.Dissemination

©Fx = max ;

360 cos 20° - 0.3(661.67) =

the student
(including work

by

protected

of the
assessing

is

solely

work provided and of

This is

work

integrity
this

the

and courses part
any
of
their