A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m=0.4kg is at rest on this surface. An impulse of 1.0Ns is applied to the block at time t=0 so that it starts moving along the x-axis with a velocity vt=v0e−t/τ, where v0 is a constant and τ=4s. The displacement of the block, in metres, at t=τ is Take e−1=0.37.
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 6.30.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
V=V0e−t/τ here V0=Jm=2.5m/s ⇒dxdt=V0e−t/τ⇒∫0xdx=V0∫0τe−t/τdt Since∫e−xdx=e−x−1⇒x=V0e-t/τ-1τ⇒x=2.5 −4 e−1−e0 ⇒x=2.5−4 0.37−1⇒x=6.30