Question

The coefficient of static friction, $\mu_{s}$, between block A of mass 2 kg and the table as shown in the figure is 0.2. What would be the maximum mass value of block B so that the two blocks do not move? The string and the pulley are assumed to be smooth and massless. (g = 10 m/s²)(a) 0.4 kg(b) 2.0 kg(c) 4.0 kg(d) 0.2 kg

Answer 1

Answer:

c) 4 kg

Explanation:

Mass of block A = 2kg (Given)

g = 10 m/s (Given)

Force applied = mg

T+ma = fu

t = uNa ( for a = 0)

T = ma+mg

t = mbg ( for a = 0)

Thus, uNa = mbg

= 0.2 ×10×2

= 4

Thus, the maximum mass value of block B so that the two blocks do not move will be 4kg

Answer 2

Answer:

0.4kg

Explanation:

For block A,

T+mAa= f

since a=0,T= f=uN ; u is coeffecient of friction

For block B,

T= mBg +mBa

since a=0, T=mBg

uN=mBg

umAg= mBg

mB=umA=0.2×2=0.4

mB= 0.4kg.