A massless cord goes horizontally from a 12 kg mass block around the rim of a su

Question

A massless cord goes horizontally from a 12 kg mass block around the rim of a suported solid disk of radius 0.50 m and mass 12 kg. The pivot is frictionless and the cord does not slip against the disk. After a half circle the cord goes back horizontally to the left. There is a 120 N force applied to the left on the upper segment of the cord. The block rests on a rough horizontal surface with a coefficient of sliding friction of 0.25. The block and disk start from rest and a 120 N force, F, is applied to the cord as shown.

After 2.0 meters displacement, what is the speed of the sliding block?

A)4.0 m/s

B)4.5 m/s

C)5.0 m/s

D)5.5 m/s

E)6.0 m/s

Explanation / Answer

B)4.5 m/s

here is the explanation

Net workdone, Wnet = F*d - Friction*d

= 120*2 - 12*9.8*0.25*2

= 181.2 J

Let I is moment of inrtia of the disk.

Let v is the speed of the block.

Net workdone = change in kinetic enrgy of the system

Wnet = 0.5*M*v^2 + 0.5*I*w^2

= 0.5*M*v^2 + 0.5*0.5*M*R^2*w^2

= 0.5*M*v^2 + 0.25*M*(R*w)^2

= 0.5*M*v^2 + 0.25*M*v^2

= 0.75*M*v^2

==> v = sqrt(Wnet/(0.75*M))

= sqrt(181.2/(0.75*12))

= 4.5 m/s

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