Hookeing

Classical Mechanics Level pending

A block of mass 4.0 kg is dropped from a height of 0.80m on to a spring of spring constant 1960 N/m. What will be the maximum distance through which the spring will be compressed? The answer must be in International System of Units.


The answer is 0.18.

Sabeel Mahmood by Sabeel Mahmood , 23, Pakistan

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Sabeel Mahmood
Mar 13, 2014

The solution is easy, start by finding potential energy, PE=mgh leading to PE= 4x9.8xx0.80 = 31.36 J Now letting the distance be x, then in case of mass-spring system, max energy = 1 2 k x 2 \frac { 1 }{ 2 } k{ x }^{ 2 }
answer is 0.1789 or 0.18m approximately.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Wrong answer Sabeer

Navin Manaswi - 7 years, 3 months ago

Log in to reply

mg(h+x) = (1/2) kx^2 answer is 0.2

Navin Manaswi - 7 years, 3 months ago

Over here you are neglecting the change in potential energy as the spring gets compressed. If the compression is x x , then the PE further decreases by m g x mgx . So, at maximum compression, the change in PE is m g ( h + x ) mg(h+x) and not m g h mgh .

As pointed out by Navin, the answer is 0.2 0.2 .

Parth Thakkar - 7 years, 2 months ago

what type of calculator do u use ?

Arijit Banerjee - 7 years, 2 months ago

Log in to reply

Possibly casio... i will change the answer... my mistake

Sabeel Mahmood - 7 years, 2 months ago

Log in to reply

no probs just edit it

Arijit Banerjee - 7 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...