Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A trolley of mass m is connected to two identical springs, each of force constant k, as shown in fig. The trolly is displaced from its equilibrium position by a distance x and released. The trolly executes simple harmonic motion of period T. After some time it comes to rest due to friction. The total energy dissipated as heat is (assume the damping force to be weak)

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

12kx2

b

kx2

c

4π2mx2T2

d

mx2T2

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

Two springs are connected in parallel equivalent spring constant is – 2K.Total energy  122Kx2=kx2The final energy of the system = 0 Hence the total energy dissipated is = kx2
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon