A person of mass 72 kg sitting on ice pushes a block of mass of 30kg on ice horizontally with a speed of 12 ms^-1. The coefficient of friction between the man and ice and between block and ice in 0.02. If g =10ms^-2, the distances between man and the block, when they come to rest is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

360 m

10 m

350 m

422.5 m

answer is D.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Velocity of man=(30x12)/72=5m/s

Final distance between Man and block will be the sum of the stopping distances of the block and man.

$s_{man} = \frac{5^{2}}{2 x 0.02 x 10} = 62.5 m$

$s_{block} = \frac{12^{2}}{2 x 0.02 x 10} = 360 m$

Final separation= 422.5m

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test