Two horizontal force $\text F1= 80 \text N$ and $\text F2= 70 \text N$ acting on a 20 kg block kept on rough horizontal surface, having coefficient of friction $= 0.4$, shown in following figure. The magnitude of the net acceleration of the block

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JKSSB JE (Civil) Official Paper (Held On: 19 Nov, 2023)

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Answer 1

Concept:

The block is acted upon by two horizontal forces:

F₁ = 80 N: F₂ = 70 N

Net Force: F_n = F₁ - F₂ = 80 - 70 = 10 N

The block is on a rough horizontal surface.

Friction force $F_f=\mu N$

where $\mu =0.4$

Force $N=mg=20\times9.81=196.2N$

So, $F_f=0.4\times196.2=78.48N$

Calculation:

Compare $F_{n e t}$ with friction:

$F_{n e t} = 10 N$
$F_{f} = 78.48 N$
Since F_net < F_f the friction will completely oppose the motion — the block will not move.
Net acceleration = 0 m/sec², because friction prevents any movement.