IIT JEE , AIEEE Forum - Physics , Mechanics

svaze

a small block of mass 100g is pressed agaist horizontal sring fixed at 1 end to compress it through 5cm Spring constant is 100N/m When released block moves horizontally till it leaves spring Where will it hit ground 2m below the spring?

risin

Here,

A=5cm or5*10^-2 m..

Wait..when it leaves the spring...it is at its maximum amplitude so velocity there is zero..?

ganesha1991

using conservation of energy
1/2mv^2 = 1/2 kx^2
v^2 = 100*5*5*10^-2*10^-2 / 0.1
= 25 * 10^-1
= 2.5

therefore
now
R = vroot 2h/g = root2.5 *root2*2/10
= root 2.5*4/10
=root100/10
= root10

krishna.gopal

Perfect answer ganesha. just one mistake. 2.5*4=10 and not hundred. So answer is 1 meter

karthik2007

Another approach would be (Though not too different from the one posted above) :

After getting the initial velocity using conservation of energy, the mass follows the path of a projectile, hence:

use

$y=xtan\theta - \frac{gx^2}{2u^2 cos^2 \theta}$

And using the fact that $\theta$ = 0, and also that y = -2, you can get your answer.

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