Question

How large must F be in the Figure shown to given the 700 gm block an acceleration of 2 30 cm / s ? The coefficient of friction between all surfaces is 0.15.

(A) 4 N 

(B) 2.18 N 

(C) 3.18 N 

(D) 6N

Answer 1

Correct option is (B) 2.18 N

Now if m₁ moves with an acceleration ‘a’ towards right; m₂ will have an acceleration of ‘a’ towards left.

[\(\because\) string constraint]

FBD of m₁;

F – f₁ – f₂ – T = m₁a   … (i)

m₁g + N₂ = N₁   … (ii)

FDB of m₂;

N₂ – m₂g = 0   … (iii)

T – f₂ = m₂a     … (iv)

f₂ = µN₂ = µm₂ g  … (v)

∴ T = m₂a + µm₂ g

T = (a + μg) m₂    … (vi)

f₁ = mN₁ = µ(m₁g + m₂g) = µg (m₁ + m₂)

∴In equation (i)

F – µg (m₁ + m₂) – µm₂g – (a + µg)m₂ = m₁a

∴ F = (m₁ + m₂)a + 3µm₂g + µm₁g

⇒ F = (m₁ + m₂)a + µg (m₁ + 3m₂)

Put a = 0.3 m/s² and m₁ = 0.7 kg, m₂ = 0.2 kg to get the value of force.

Hence, we get F = 2.18 N.

Answer 2

Correct answer is (B) 2.18 N