A block of mass 5 kg moves along the x-direction subject to the force F = (-20x + 10)N,

Question

A block of mass 5 kg moves along the x-direction subject to the force \(F=(-20 x+10) \mathrm{N}\), with the value of x in metre. At time \(t=0 \mathrm{~s}\), it is at rest at position \(x=1 \mathrm{~m}\). The position and momentum of the block at \(t=(\pi / 4) \mathrm{s}\) are

(A) \(-0.5 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

(B) \(0.5 \mathrm{~m}, 0 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

(C) \(0.5 \mathrm{~m},-5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

(D) \(-1 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

Answer 1

Correct option is (C) \(0.5 \mathrm{~m},-5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

\(F = -20 \left(x- \frac {1}{2}\right)= -20 X\)  \(\left(X = x - \frac {1}{2}\right)\)

\(\therefore \) Particle will perform SHM about \(x = \frac {1}{2}\) with

\(\omega = 2\,rad /sec \Rightarrow T = \pi \,sec.\)

\(\therefore \) Phase covered in \(t = \frac {\pi}{4}\) second = \(90 ^\circ\)

Given particle is at rest at x = 1m ⇒ x = 1 is extreme position.

\(\therefore \) In \(\frac {\pi}{4}\) sec, it will be at equilibrium

\(\therefore \) x = 0.5 m and momentum \(= m \,\omega A = 5 \times 2 \times 0.5 = 5\) kg m/s 

Direction will be towards -ve x.

Comments

Ok.........