A spring 40 , mm long is stretched by the app | vedclass

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

Question

(D) The spring constant $k$ is given by Hooke's Law: $k = \frac{F}{x}$.Given $F = 10 \, N$ and $x = 1 \, mm = 1 	imes 10^{-3} \, m$.So,$k = \frac{10}

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The spring constant $k$ is given by Hooke's Law: $k = \frac{F}{x}$.Given $F = 10 \, N$ and $x = 1 \, mm = 1 	imes 10^{-3} \, m$.So,$k = \frac{10}{1 	imes 10^{-3}} = 10^4 \, N/m$.The work done $W$ in stretching the spring by a displacement $x = 40 \, mm = 40 	imes 10^{-3} \, m$ is given by $W = \frac{1}{2} k x^2$.Substituting the values: $W = \frac{1}{2} 	imes 10^4 	imes (40 	imes 10^{-3})^2$.$W = \frac{1}{2} 	imes 10^4 	imes (1600 	imes 10^{-6}) = \frac{1}{2} 	imes 10^4 	imes 1.6 	imes 10^{-3} = 0.5 	imes 16 = 8 \, J$.

$A$ spring $40 \, mm$ long is stretched by the application of a force. If $10 \, N$ force is required to stretch the spring by $1 \, mm$,then the work done in stretching the spring by $40 \, mm$ is ............. $J$.

Similar Questions

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

Show that the law of conservation of mechanical energy is obeyed by pulling or compressing the block tied at the end of a spring.

When equal forces are applied to two springs with spring constants $1500 \, N/m$ and $3000 \, N/m$,what is the ratio of their stored potential energies?

$A$ block of mass $M$ moving on a frictionless horizontal surface collides with a spring of spring constant $K$ and compresses it by a length $L$. The maximum momentum of the block during the collision process is

Does the energy stored in a spring change when it is stretched or compressed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module