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4.1 , newtons laws of motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics4.1 , newtons laws of motion1 Mark
MCQ
A frictionless cart $A$ of mass $100\  kg$ carries other two frictionless carts $B$ and $C$ having masses $8\  kg$ and $4\  kg$ respectively connected by a string passing over a pulley as shown in the figure. What horizontal force $F$ must be applied on the cart so that smaller cart do not move relative to  it .......... $N$
  • A
    $150$
  • B
    $340$
  • ✓
    $560$
  • D
    $630$
✓

Answer

Correct option: C.
$560$
c
Let $a$ be the acceleration of the system in the right direction.

since the smaller carts do not move relative to the larger one, the equation of motion of the whole system,

$\left(m_{A}+m_{B}+m_{C}\right) a=F$

$(100+8+4) a=F$

$\Longrightarrow F=112 a$

For block $\mathrm{B}, T=m_{B} a=8 a \quad(...1)$

For block $\mathrm{C}, T=m_{C} g=4 g$         $...(2)$

From $(1)$ and $(2),$ we get $8 a=4 g$

Thus $a=0.5 g=5 \quad\left(\text { as } g=10 m / s^{2}\right)$

$\therefore F=112 \times 5=560 N$

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