Jim Guinn’s PHYS1111 Assignment #9
Due Wednesday, Feb. 24, 2010
1. A 1.00kg mass is attached to a wall with a string as shown in the diagram.
It is resting on a
2.00kg mass.
The 2.00kg mass is pulled to the right with a force of 20.0N .
There is a
coefficient of kinetic friction of 0.400 between the two masses and between the 2.00kg mass
and the floor.
a.)
What is the tension is the string attached to the 1.00kg mass?
The forces acting on the 1.00kg block are
F
y
= n – W = ma
y
= 0 , so n = W = mg = (1.00kg) (9.80m/s
2
) =
9.80N .
F
x
= f
k
– T = ma
x
= 0 , so T = f
k
=
μ
k
n = (0.400) (9.80N) =
T = 3.92N .
b.) What is the acceleration of the 2.00kg mass?
The forces acting on the 2.00kg block are
F
y
= n
floor
– W – n
1kg
= ma
y
= 0 , so
n
floor
= W + n
1kg
= (2.00kg)(9.80m/s
2
) + 9.80N = 29.4N .
F
x
= 20.0N – f
floor
– f
1kg
= 20.0N –
μ
k
n
floor
– 3.92N = 20.0N – (0.400)(29.4N) – 3.92N =
= 4.32N = ma
x
= (2.00kg)a
x
so
a
x
= 2.16m/s
2
.
2.
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 Spring '11
 lomant
 Force, Mass, Nfloor

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