Question

a block of 10kg mass is placwd on a table and a block of 20kg mass is suspended vertically by means of a string passing over a pulley if coefficient of kinetic friction is 0.4 the resulting acceleration of the system is​

Answer 1

Hii Abhishek,

◆ Answer -

a = 5.228 m/s^2

● Explaination -

# Given -

m1 = 10 kg

m2 = 20 kg

μ = 0.4

# Solution -

Let T be the tension in the string and a be resultant acceleration.

Looking at the figure,

m2.a = m2.g - T

T = m2.g - m2.a ...(1)

Also,

m1.a = T - m1.g

T = m1.a + μ.m1.g ...(2)

From (1) & (2),

m2.g - m2.a = m1.a + μ.m1.g

m2.g - μ.m1.g = m1.a + m2.a

g (m2-μ.m1) = a (m1+m2)

a = g (m2-μ.m1) / (m1+m2)

Substituting valuez,

a = 9.8 × (20 - 0.4×10) / (10 + 20)

a = 9.8 × (20-4) / 30

a = 5.228 m/s^2

Therefore, resulting acceleration of the system is 5.228 m/s^2.

Best luck dear...