Download Ffr = ma

1. A free-body diagram for the box is shown, assuming that it is moving to the right. The
“push” is not shown on the free-body diagram because as soon as the box moves away from
the source of the pushing force, the push is no longer applied to the box. It is apparent from
the diagram that FN = mg for the vertical direction. We write Newton’s 2nd law for the
horizontal direction, with positive to the right, to find the acceleration of the box.
∑F
x
− Ffr = ma
ma = − µ k FN = − µ k mg
→
(
→
)
a = − µ k g = −0.2 9.8 m s 2 = −1.96 m s 2
The initial speed is 4.0 m/s, and the final speed will be 0. Then,
v − v = 2a (x − x0 )
2
2
0
v 2 − v02 0 − (4.0 m s )
x − x0 =
=
= 4.1 m
2a
2 −1.96 m s 2
2
→
(
)
2. a) Consider the free-body diagram for the carton on the surface. There is no motion in the y
direction and thus and hence no acceleration in the y direction. Write Newton’s 2nd law for
both directions.
x
= FN − mg cos θ = 0
x
= mg sin θ − Ffr = ma
∑F
∑F
→
FN = mg cos θ
ma = mg sin θ − µ k FN = mg sin θ − µ k mg cos θ
a = g (sin θ − µ k cos θ )
(
)
= 9.80 m s 2 (sin 22.0º − 0.12 cos 22.0º ) = 2.58 = 2.6 m s 2
(b) Now use Eq. 2-11c, with an initial velocity of 0, to find the final velocity.
v 2 − v02 = 2a (x − x 0 ) →
(
3. We derive two expressions for acceleration – one from the
kinematics, and one from the dynamics. With a starting speed of vo
up the plane and a final speed of zero, we have
v 2f − v o2 = 2 a ( x − x 0 ) → a =
)
v = 2a (x − x 0 ) = 2 2.58 m s 2 (9.30 m ) = 6.9 m s
− v o2
− v o2
=
2 (x − x 0 )
2d
Write Newton’s 2nd law for both the x and y directions. Note that
the net force in the y-direction is zero, since the block does not
accelerate in the y-direction.
∑F
∑F
y
= F N − mg cos θ = 0 → F N = mg cos θ
x
= − mg sin θ − F fr = ma
→ a =
− mg sin θ − F fr
m
Now equate the two expressions for the acceleration, substitute in the relationship between
the frictional force and the normal force, and solve for the coefficient of friction.
− mg sin θ − Ffr
− v o2
=
m
2d
a =
µk
→
mg sin θ + µ k mg cos θ
v2
= o
m
2d
→
v o2
=
− tan θ
2 gd cos θ
4. We have:
0.250 ;
either when the block slides up or down the wall#
$ 30.0& ;
'( 39.0 *
We want to calculate + ,+-.,.
(a) We have the following free-body diagram for when the block slides up the wall:
>
0.2
θ
+-.
+-.
90& 6 $
θ
'(.
?
0.9:
There is no motion in the x direction:
/ 01 02 cos180& # 5 + cos90& 6 $# 0
In the y-direction:
02 + cos90& 6 $#
/ 07 + sin90& 6 $# 5 '( sin270& # 5 09: sin270& # 0
;<=>#
+
'( 5 09:
sin90& 6 $#
Now,
09: 02 + cos90& 6 $#
Then,
+
+
'( 5 + cos90& 6 $#
sin90& 6 $#
39.0* 5 0.250#+# cos60& #
sin60& #
+sin60& # 6 0.250 A cos60& ## 39.0*
+
39.0*
0.74
+ 52.6 *
(b) We have the following free-body diagram for when the block slides down the wall:
0.9:
0.2
θ
+-.
+-.
90& 6 $
θ
'(.
There is no motion in the x direction:
/ 01 02 cos180& # 5 + cos90& 6 $# 0
In the y-direction:
>
02 + cos90& 6 $#
?
/ 07 + sin90& 6 $# 5 '( sin270& # 5 09: sin90& # 0
'( 6 09:
+
sin90& 6 $#
;<=>#
Now,
Then,
09: 02 + cos90& 6 $#
+
'( 6 + cos90& 6 $#
sin90& 6 $#
39.0* 6 0.250#+# cos60& #
+
sin60& #
+sin60& # 5 0.250 A cos60& ## 39.0*
+
39.0*
0.99
+ 39.4 *