Two identical springs have the same force con | vedclass

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

Question

(D) Given: Force constant $k = 73.5 \,Nm^{-1}$,Mass $m = 5 \,kg$,Acceleration due to gravity $g = 9.8 \,ms^{-2}$.The weight acting on the springs is $

Vedclass · Accepted Answer

$\frac{1}{3} \,m, \frac{4}{3} \,m, \frac{2}{3} \,m$

Vedclass · Answer

(D) Given: Force constant $k = 73.5 \,Nm^{-1}$,Mass $m = 5 \,kg$,Acceleration due to gravity $g = 9.8 \,ms^{-2}$.The weight acting on the springs is $F = mg = 5 	imes 9.8 = 49 \,N$.In Figure-$1$,the two springs are in parallel. The equivalent spring constant is $k_{eq} = k + k = 2k$.The elongation $x_1$ is given by $F = k_{eq} x_1 \implies 49 = (2 	imes 73.5) x_1 \implies x_1 = \frac{49}{147} = \frac{1}{3} \,m$.In Figure-$2$,the two springs are in series. The equivalent spring constant is $k_{eq} = \frac{k 	imes k}{k + k} = \frac{k}{2}$.The elongation $x_2$ is given by $F = k_{eq} x_2 \implies 49 = (\frac{73.5}{2}) x_2 \implies x_2 = \frac{49 	imes 2}{73.5} = \frac{98}{73.5} = \frac{4}{3} \,m$.In Figure-$3$,there is only one spring. The elongation $x_3$ is given by $F = k x_3 \implies 49 = 73.5 x_3 \implies x_3 = \frac{49}{73.5} = \frac{2}{3} \,m$.Thus,the elongations are $\frac{1}{3} \,m, \frac{4}{3} \,m, \frac{2}{3} \,m$.

Two identical springs have the same force constant $73.5 \,Nm^{-1}$. The elongation produced in each spring in the three cases shown in Figure-$1$,Figure-$2$,and Figure-$3$ are (given $g=9.8 \,ms^{-2}$):

Similar Questions

$A$ spring with $10$ coils has spring constant $k$. It is exactly cut into two halves,then each of these new springs will have a spring constant

$A$ spring has a spring constant $k$ and original length $l$. If it is cut into pieces in the ratio $\alpha : \beta : \gamma$,find the spring constant of each piece in terms of the original spring constant $k$ (where $\alpha, \beta$,and $\gamma$ are integers).

$A$ long spring is stretched by $2 \ cm$ and its potential energy is $U$. If the spring is stretched by $10 \ cm$,its potential energy will be

Initially,the spring is at its natural length and both blocks are at rest. Determine the maximum extension in the spring. Given $k = 20 \ N/m$,$m_1 = 0.5 \ kg$,$m_2 = 1 \ kg$,and $F = 1 \ N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module