A light elastic spring is placed on a massles | vedclass

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

Question

(C) According to the law of conservation of energy,the potential energy of the particle at height $h$ plus the compression $x$ is converted into the e

Vedclass · Accepted Answer

$3.96$

Vedclass · Answer

(C) According to the law of conservation of energy,the potential energy of the particle at height $h$ plus the compression $x$ is converted into the elastic potential energy of the spring:$mg(h + x) = \frac{1}{2} K x^2$Case $I$: $m = 0.1 \ kg$,$h_1 = 0.24 \ m$,$x_1 = 0.01 \ m$$mg(0.24 + 0.01) = \frac{1}{2} K (0.01)^2$$mg(0.25) = \frac{1}{2} K (0.0001) \quad ... (1)$Case $II$: $m = 0.1 \ kg$,$h_2 = h$,$x_2 = 0.04 \ m$$mg(h + 0.04) = \frac{1}{2} K (0.04)^2$$mg(h + 0.04) = \frac{1}{2} K (0.0016) \quad ... (2)$Dividing equation $(2)$ by $(1)$:$\frac{h + 0.04}{0.25} = \frac{0.0016}{0.0001} = 16$$h + 0.04 = 16 	imes 0.25$$h + 0.04 = 4$$h = 4 - 0.04 = 3.96 \ m$

Similar Questions

$A$ car of mass $1000 \,kg$ having a velocity of $10 \,ms^{-1}$ collides with a horizontally mounted spring. If the spring constant is $4000 \,Nm^{-1}$, then the maximum compression of the spring is: (in $\,m$)

The potential energy of a long spring when stretched by $2 \ cm$ is $U$. If the spring is stretched by $8 \ cm$,the potential energy stored in it is:

Two springs have their force constants as $k_1$ and $k_2$ $(k_1 > k_2)$. When they are stretched by the same force $F$,which of the following statements is true regarding the work done?

Vedclass Test Series

Exam Paper Generator

Online Exam Module