A mass of 2 kg,initially at a height of 1.2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the compression of the spring be $x$ (in meters). According to the law of conservation of mechanical energy,the total potential energy lost by

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) Let the compression of the spring be $x$ (in meters). According to the law of conservation of mechanical energy,the total potential energy lost by the mass equals the potential energy gained by the spring. The total height fallen by the mass is $(h + x)$,where $h = 1.2 \ m$. So,$mg(h + x) = \frac{1}{2} kx^2$. Substituting the given values: $2 	imes 10 	imes (1.2 + x) = \frac{1}{2} 	imes (2 	imes 10^4) 	imes x^2$. $20(1.2 + x) = 10^4 x^2$. $24 + 20x = 10000x^2$. $10000x^2 - 20x - 24 = 0$. Dividing by $4$: $2500x^2 - 5x - 6 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{5 \pm \sqrt{(-5)^2 - 4(2500)(-6)}}{2(2500)} = \frac{5 \pm \sqrt{25 + 60000}}{5000} = \frac{5 \pm \sqrt{60025}}{5000} = \frac{5 \pm 245}{5000}$. Since $x > 0$,$x = \frac{250}{5000} = 0.05 \ m$. Converting to $mm$: $0.05 \ m = 50 \ mm$.

$A$ mass of $2 \ kg$,initially at a height of $1.2 \ m$ above an uncompressed spring with spring constant $2 \times 10^4 \ N/m$,is released from rest to fall on the spring. Taking the acceleration due to gravity as $10 \ m/s^2$ and neglecting air resistance,the compression of the spring in $mm$ is:

Similar Questions

$A$ pendulum bob has a speed of $3 \, m/s$ at its lowest position. The pendulum is $50 \, cm$ long. The speed of the bob, when the string makes an angle of $60^{\circ}$ to the vertical, will be $....... \, m/s$ $(g = 10 \, m/s^2)$.

$A$ particle of mass $500 \,g$ is moving in a straight line with velocity $v = b x^{5/2}$. The work done by the net force during its displacement from $x = 0$ to $x = 4 \,m$ is ................... $J$ (Take $b = 0.25 \,m^{-3/2} s^{-1}$)

$A$ body of mass $1 \text{ kg}$ moves along a straight line with a velocity $v = 2x^2$. The work done by the body during displacement from $x = 0$ to $5 \text{ m}$ is . . . . . . $J$.

What does the law of conservation of energy state?

Vedclass Test Series

Exam Paper Generator

Online Exam Module