A block of mass m is lying on a wedge having inclination angle α=tan−115. Wedge is moving with a constant acceleration a=2ms−2. The minimum value of coefficient of friction μ so that m remains stationary w.r.t. wedge 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
a
2/9
b
5/12
c
1/5
d
2/5
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
FBD of m in frame of wedge, N=mg cos α−ma sin αNow f=μN=macosα+mgsinα μ=acosα+gsinαgcosα−asinα=a+gtanαg−atanα=512
Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!
A block of mass m is lying on a wedge having inclination angle α=tan−115. Wedge is moving with a constant acceleration a=2ms−2. The minimum value of coefficient of friction μ so that m remains stationary w.r.t. wedge is