Two blocks of equal mass m are connected by an unstretched spring and the system is kept at rest on a frictionless horizontal surface. At =0, a constant force F is applied on the first block pulling it away from the other as shown in figure.

If the extension of the spring is $x_{0}$ at time t, then the displacement of the first block in time t 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

$\frac{1}{2} (\frac{F t^{2}}{2 m} + x_{0})$

$- \frac{1}{2} (\frac{F t^{2}}{4 m} + x_{0})$

$\frac{1}{2} (\frac{F t^{2}}{2 m} - x_{0})$

$(\frac{F t^{2}}{2 m} + x_{0})$

answer is A.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

${\vec{S}}_{b l o c k w r t g} = {\vec{S}}_{w r t c m} + {\vec{S}}_{c m, g} = \frac{x_{0}}{2} + \frac{1}{2} \times a_{c m} \times t^{2}$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

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

Take Free Test