Question

a block of 9 kg moving with velocity of 4 m/s on frictionless surface the block comes into rest by compressing the springs if springs constant is 4×10^4 N/kg then find compression in the springs ​

Answer 1

Given A block of 9 kg moving with velocity of 4 m/s on friction less surface the block comes into rest by compressing the springs if springs constant is 4×10^4 N/m then find compression in the springs.

Energy possessed by a body by virtue of its position is potential energy and energy possessed by a body by virtue of its motion is kinetic energy.

From law of conservation of energy P.E = K.E

                                                  1/2 p x^2 = 1/2 m v^2

                                               x^2 = mv^2 / p

                                              x^2 = 9 x 4 / 40000

                                           x = √9 x 4 / 40000

                                         x = 3 x 2 / 200

                                         x = 3/100 m

Answer 2

Answer:

A 9 kg block compresses an ideal spring  and comes to rest. If the force constant of spring is $4*10^{4}$ N/m, then spring compressed distance is $22*10^{-4}m$

Explanation:

• The restoring force on a spring of spring constant (K) when elongated to a distance (x) is given as $F=-Kx$

• Due to the mass (m) attached to the spring the gravitational force on the mass is  $F=mg$

        where $g=9.8m/s^{2}$(acceleration due to gravity)

• at equilibrium condition  both the forces are equal, that is $-Kx=mg$

From the question, we have

mass attached $m=9kg$

spring constant $K=4*10^{4} N/m$

stretched length $x=mg/K$

put the given values

⇒$x=\frac{9*9.8}{4*10^{4} } =22*10^{-4} m$