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PHYSICS
Static friction Force
F
a1
=P
c1
F
a1
=490
[N]
* 0,02
F
a1
≅
9,8
N
Kinetic friction
F
a2
=P
c2
F
a2
=490
[N]
* 0,01
F
a2
≅
4,9
N
Acceleration
a
=
2
S
2 * 2
[m]
4
[m]
a
=
0,16
m
⁄
s
2
t
2
5
[s
2
]
25
[s
2
]
To be able tomove the body with an acceleration equal to 0,16
[m
⁄
s
2
]
it is necessary for the resultant horizontal
Force to be:
F
Ro
=m * a
50 [Kg] *
0,16
[N / s
2
]
F
Ro
=
8
N
The horizontal driving Force appliedwill be:
F
mo
= F
Ro
+
F
a2
8 [N]
+4,9
[N]
F
mo
=
12,9
N
The obvious difference in the Forces between
F
m
ν
and F
mo
is due to the vertical equilibrium condition, the Force
F
e
must oppose the Force of gravity acting downwards on the body, this is not a factor in the horizontal condition,
as the supporting plane provides an equal and opposite Force to support theweight of the body.
Velocity
ν
= a * t
0,16
[m
/
s
2
]
* 5
[s]
ν=
0,80
m / s
Thework required tomove the body is:
L= F
mo
* S
12,9
[N]
* 2
[m]
L
=
25,8
J
A part of it will be dissipated to overcome the friction Force and a part will be transferred to the body in the form
of kinetic energy:
E
=
1
m
*
ν
2
1
50
[Kg]
*
0,8
2
[m
/
s
2
]
E
=
16
J
2
2
Assuming that the body is awheel of radius
r
equal to 500
mm
, the rolling friction Forcewould be:
F
mo
=
P b
490
[N]
*
0,002
[m]
F
a
ν
≅
1,96
N
r
0,5
[m]
And the horizontal driving Force:
F
mo
= F
Ro
+ F
a
ν
8
[N]
+1,96
[N]
F
mo
=
9,96
N
1
33
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