An object of mass m falls on to a spring of constant k from height h. The spring undergoes compression by a length x. The maximum compression x is given by the equation:

Question

An object of mass m falls on to a spring of constant k from height h. The spring undergoes compression by a length x. The maximum compression x is given by the equation: (A) m g h=(1/2) k x2 (B) m g(h+x)=(1/2) k x2 (C)  m g(h+x)=-kx  (D)  m g h=-kx

Tardigrade Admin · Accepted Answer

Energy is conserved. From law of conservation of energy potential energy at height h= energy stored in compressed spring. ∴ m g h+m g x=(1/2) k x2 ⇒ m g(h+x)=(1/2) k x2

Q. An object of mass $m$ falls on to a spring of constant $k$ from height $h .$ The spring undergoes compression by a length $x .$ The maximum compression $x$ is given by the equation:

$m g h = \frac{1}{2} k x^{2}$

$m g (h + x) = \frac{1}{2} k x^{2}$

$m g (h + x) = - k x$

$m g h = - k x$

Energy is conserved. From law of conservation of energy potential energy at height $h =$ energy stored in compressed spring.
$∴ m g h + m gx = \frac{1}{2} k x^{2}$
$\Rightarrow m g (h + x) = \frac{1}{2} k x^{2}$

Q. An object of mass m falls on to a spring of constant k from height h. The spring undergoes compression by a length x. The maximum compression x is given by the equation:

mgh=21​kx2

mg(h+x)=21​kx2

mg(h+x)=−kx

mgh=−kx