A man applies a force F on a spring-block system shown, towards right when the block is at rest and spring is relaxed. If F is constant, then

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

When the block passes through equilibrium position, the energy stored in the spring and the kinetic energy of the block respectively, are $\frac{F^{2}}{2 k} a n d \frac{F^{2}}{2 k}$

The block does not oscillate

the time period of the block is $2 π \sqrt{\frac{M}{k}}$

the amplitude is $\frac{F}{k}$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

At extreme position, $F \cdot 2 A = \frac{1}{2} k (2 A)^{2}$

$\Rightarrow A = \frac{F}{k}$

At mean position, work done by $F = \frac{1}{2} k A^{2} + \frac{1}{2} m v^{2}$

$∴ F \cdot A = \frac{1}{2} k A^{2} + (KE)_{mean position} F \cdot \frac{F}{k} - \frac{1}{2} k {(\frac{F}{k})}^{2} = (KE)_{mean position} (KE)_{mean position} = \frac{F^{2}}{2 k} ∴ (U)_{mean position} = \frac{F^{2}}{k} - \frac{F^{2}}{2 k} = \frac{F^{2}}{2 k}$