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AIIMS 201826 May
Exam
AIIMS
Year
2018
Date
26 May
Subject
Physics
Chapter
Laws of Motion

AIIMS 2018 · 26 May · Physics · Laws of Motion

The rear side of a truck is open and a box of mass 20 ~kg is placed...

The rear side of a truck is open and a box of mass 20 kg20 \mathrm{~kg} is placed on the truck 4 m4 \mathrm{~m} away from the open end. If μ\mu is 0.15 and gg is 10 m s−210 \mathrm{~m} \mathrm{~s}^{-2} and the truck starts from rest with an acceleration of 2 m s−22 \mathrm{~m} \mathrm{~s}^{-2} on a straight road, then the box will fall off the truck when it is at a distance of xx metre from the starting point. The value of xx isThe rear side of a truck is open and a box of mass 20 ~kg is placed on the truck 4 ~m away from the open end. If μ is

Options

  1. A4 m4 \mathrm{~m}
  2. B8 m8 \mathrm{~m}
  3. C16 m16 \mathrm{~m}
  4. D32 m32 \mathrm{~m}

Correct answer

C. 16 m16 \mathrm{~m}

Step-by-step solution

a=2 m s−2,N=mga=2 \mathrm{~m} \mathrm{~s}^{-2}, N=m g For the box's motion w.r.t. truck, ma−μN=maBT⇒(ma−μmg)=maBT⇒aBT=a−μg\begin{aligned} & m a-\mu N=m a_{B T} \Rightarrow(m a-\mu m g)=m a_{B T} \\ & \Rightarrow a_{B T}=a-\mu g\end{aligned}=2 m s−2−(0.15)(10 m s−2)=0.5 m s−2\begin{aligned} & =2 \mathrm{~m} \mathrm{~s}^{-2}-(0.15)\left(10 \mathrm{~m} \mathrm{~s}^{-2}\right) \\ & =0.5 \mathrm{~m} \mathrm{~s}^{-2}\end{aligned} sBT=uBT⋅t+12aBT⋅t24=0+12(0.5)t2⇒t2=16⇒t=4 s.\begin{aligned} & s_{B T}=u_{B T} \cdot t+\frac{1}{2} a_{B T} \cdot t^2 \\ & 4=0+\frac{1}{2}(0.5) t^2 \Rightarrow t^2=16 \\ & \Rightarrow t=4 \mathrm{~s} .\end{aligned}AfterAfter4 \mathrm{~s},theboxwillfalloffthetruck.During, the box will fall off the truck. During4 \mathrm{~s},thetrucktravels,, the truck travels,x=\frac{1}{2} a t^2=\frac{1}{2}(2)(4)^2=16 \mathrm{~m}$. The rear side of a truck is open and a box of mass 20 ~kg is placed on the truck 4 ~m away from the open end. If μ is
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