A block of mass minitially at rest is dropped from a height h on to a spring of force constant k. the maximum compression in the spring is x then

Question

A block of mass minitially at rest is dropped from a height h on to a spring of force constant k. the maximum compression in the spring is x then (A) m g h=(1/2) k x2 (B) m g(h+x)=(1/2) k x2 (C) m g h=(1/2) k(x+h)2 (D) m g(h+x)=(1/2) k(x+h)2

Tardigrade Admin · Accepted Answer

Change in gravitational potential energy = Elastic potential energy stored in compressed spring ⇒ m g(h +x)=(1/2) k x2

Q. A block of mass minitially at rest is dropped from a height $h$ on to a spring of force constant $k$ . the maximum compression in the spring is $x$ then

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

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

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

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

Change in gravitational potential energy = Elastic potential energy stored in compressed spring
$\Rightarrow m g (h + x) = \frac{1}{2} k x^{2}$

Q. A block of mass minitially at rest is dropped from a height h on to a spring of force constant k. the maximum compression in the spring is x then

mgh=21​kx2

mg(h+x)=21​kx2

mgh=21​k(x+h)2

mg(h+x)=21​k(x+h)2