Question #99672

A lump of metal of mass 5.0 kg is lying on a table top. It is tied to a string which runs over a pulley at the edge of the table. The other end of the string holds a 2.0 kg mass hanging down. Friction on the table is µ = 0.2 a) How much friction between the larger mass and the table? b) Is there any friction on the smaller mass? c) How much weight makes the system move? d) How much mass is moving? e) How much force is available to make the system move? f) How much acceleration can occur? G) How much force acts through the string to make the larger mass move

Expert's answer

a) How much friction between the larger mass and the table

f=μ×M×gf=\mu\times M\times g

f=0.2x5x9.81=9.81N

b) Is there any friction on the smaller mass?

Ans: No: Friction is only acting of two surface in contact.Here Air friction is neglected

c) How much weight makes the system move?

a = (m - μM)g/(m + M)

a=(2-0.2*5)*9.81/(2+5)

a=1.401m/s2

F=ma=2*1.401=2.80N

Weight required greater than force of friction i.e greater than 9.81 N

d) How much mass is moving?

No

e) How much force is available to make the system move?

a = (m - μM)g/(m + M)

a=(2-0.2*5)*9.81/(2+5)

a=1.401m/s2

F=ma=2*1.401=2.80N

F=ma=2*1.401=2.80N

f) How much acceleration can occur?

acceleration a=1.401m/s2

G) How much force acts through the string to make the larger mass move

T - f = Ma.

T=Ma+f

T=5x1.401+9.81

T=7.007+9.81=16.81N


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