Question

A block of mass $100\text{kg}$ is attached to the back of a truck with the help of a spring. The frictional coefficient between the truck and the block is $0.2$ (take ${\mu }_s={\mu }_k)$. Acceleration of the truck is $5{\text{m/s}}^2$ and the spring elongates by a distance of $1\text{cm}$. If the block is just about to slip, what is the force constant of the spring? Take $g=10{\text{m/s}}^2$.

• A. $300\text{N/m}$
• B. $30000\text{N/m}$
• C. $3000\text{N/m}$
• D. $1000\text{N/m}$

Answer 1

The correct option is B $30000\text{N/m}$

As the truck accelerates, spring will start to elongate so that it can provide the required force for acceleration in the $x-$direction.

Also, friction will act at its limiting value $f_{max}$ to support the spring force, in order to avoid slipping between block and truck.

Solving from accelerated frame will ease the problem, since block will be at rest in that frame, but we will apply a $F_{\text{pseudo}}$ in opposite direction on block in the FBD shown below:


$F_{\text{pseudo}}=ma=100\times 5=500\text{N}$

Spring force $=kx$, where $x=0.01\text{m}=$ elongation in the spring

From the FBD, applying equilibrium condition in horizontal and vertical directions:

$\Rightarrow kx+f_{max}-F_{\text{pseudo}}=\mathrm{0.......}i$
$N=mg..............ii$
and $f_{max}={\mu }_{s}N={\mu }_{s}mg=200\text{N}.....iii$

Solving Eq. $i$ , $ii$ , $iii$ gives:

$\Rightarrow k\times \left(0.01\right)+200-500=0$

$\therefore k=30000\text{N/m}$