Question

ब्लॉक मूवी 90-theta 30 डिग्री विद ए वेलोसिटी 5 मीटर पर सेकंड स्टॉप अफरोज ओपन 5 सेकंड व्हाट इज द वैल्यू ऑफ फॉर पेशेंट ऑफ फ्रिक्शन​

Answer 1

Answer:

Explanation:

2kg block has relative motion towards right.Therefore , maximum friction will acts on it towards left

f=μN=(0.4)(1)(10)=4N

a1=4/1=4m/s^2

a2=−4/2=2m/s^2

(b) Relative motion between them will stop when ,

v1=v1 ...(i)

∴u1+a1t=u2+a2t ...(ii)

∴2+4t=8−2t ...(iv) or t=1s

(b) Substituting value of 't' in Eq. (ii),either on RHS or on LHS common velocity =6m/s

( c) s1=u1t+12a1t2

=(2)(1)+12(4)(1)2=4m

s2=u2t+12a2t2

=(8)(1)+12(−2)(1)2=7m