A boy of mass 4 kg is standing on a piece of wood having mass

Question:

A boy of mass $4 \mathrm{~kg}$ is standing on a piece of wood having mass $5 \mathrm{~kg}$. If the coefficient of friction between the wood and the floor is $0.5$, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is N.(Round off to the Nearest Integer)

$\left[\right.$ Take $\left.g=10 \mathrm{~ms}^{-2}\right]$

Solution:

$\mathrm{N}+\mathrm{T}=90$

$\mathrm{~T}=\mu \mathrm{N}=0.5(90-\mathrm{T})$

$1.5 \mathrm{~T}=45$

$\mathrm{~T}=30$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now