How Forces Affect Motion: CBSE Class 9 Physics practice questions

D. 490 N

Total mass $= 20 + 15 + 15 = 50\ \text{kg}$. Weight $= mg = 50 \times 9.8 = 490\ \text{N}$ downwards. To keep the barbell steady the forces must be balanced, so she pushes 490 N upwards. 343 N counts only one plate, 196 N counts only the bar, and 50 N is the mass, not a force.

C. $120\ \text{m s}^{-2}$

$50\ \text{g} = 0.05\ \text{kg}$, so $a = \dfrac{6}{0.05} = 120\ \text{m s}^{-2}$. By Newton's third law the bullet pushes the gun back with the same 6 N, giving the gun only $\dfrac{6}{4} = 1.5\ \text{m s}^{-2}$ (that option is the gun's value). Equal forces, unequal masses, so unequal accelerations. 0.12 comes from using 50 kg, and 0.3 from multiplying force by mass.

C. 2400 N towards the west

$u = 12\ \text{m s}^{-1}$, $v = 0$, $t = 16 - 10 = 6\ \text{s}$, so $a = \dfrac{0 - 12}{6} = -2\ \text{m s}^{-2}$. $$F = ma = 1200 \times (-2) = -2400\ \text{N}$$ The minus sign means the force acts opposite to the motion, i.e. towards the west. 'East' is the sign slip, 1440 N uses $t = 10$ s read off the axis instead of the 6 s interval, and the car is slowing down, so the force is not zero.