Question

Objects with masses m1 = 16.0 kg
and m2 = 9.0 kg
are connected by a light string that passes over a frictionless pulley as in the figure below. If, when the system starts from rest, m2 falls 1.00 m in 1.46 s, determine the coefficient of kinetic friction between m1 and the table.
An illustration shows a rectangular block of mass m1 on the horizontal surface of a table. This is connected to another rectangular block of mass m2 by a cord that runs over a pulley placed diagonally at the corner of the horizontal surface of the table. The rectangular block of mass m2 is suspended vertically by the side of the tabletop.
(b)
What If? What would the minimum value of the coefficient of static friction need to be for the system not to move when released from rest?

Answer 1

Answer:

The acceleration of the system is found from,

Δy=v

y

t+

2

1

a

y

t

2

since v

yi

=0, we obtain

a=

t

2

2Δy

=

(1.20s)

2

2(1.00m)

=1.39m/s

2

Using the free-body diagram for m

2

, Newton's second law gives,

∑F

y2

=m

2

a:

            m

2

g−T=m

2

a

            T=m

2

(g−a)

             =(5.00kg)(9.80m/s

2

−1.39m/s

2

)

             =42.1N

Then, applying Newton’s second law to the horizontal motion of m

1

:

∑F

x1

=m

1

a:

            T−f=m

1

a

            f=T−m

1

a

             =42.1N−(10.0kg)(1.39m/s

2

)=28.2N

Since n=m

1

g=98.0N, we have

     μ

k

=

n

f

=

98.0N

28.2N

=0.288