Pumping rubber

To solve this problem you must use Hooke's law F=kx where k is spring constant (400N/m) and x is to be found. We can work out the force by using f=ma, so 10*9.8=98N. Re-arrange the first equation to get F/k=x so 98/400=0.245m

spring constant=force/elongation; k=f/e; k=mg/e; 400=10*9.8/e; e=98/400; e=0.245

If we pull the rubber band through a distance of $x$ metres , the restoring force produced = $-kx$ , where $k$ is the spring constant . By the given condition , $-kx=mg$ . Therefore , $x=\frac{-mg}{k}$ . Substituting $m=10kg$ , $k=400N/m$ and $g=-9.8m/s^2$ , we get $x=0.245$ m .

$F=k.x$ so, $x=\frac{F}{k}$ $x=\frac{9.8x10}{400} = 0.245 meters$

HERE mg=kx..so x=mg/k=10*9.8/400=.245

the gravity on 10kg mass is 98N 400 N/m /98 N =0.245

$F=kx$ where $F$ is $Force$ , $k$ is $Spring$ $Constant$ and $x$ is $distance$

$F=10*9.8=98$ and $k=400$

$98=400x$

$x=\frac{98}{400}=\boxed{0.245}$

spring force formula is : F=kx where k is the spring constant and x is the compression distance spring force also equal to gravitational force on a 10 kg weight ( given)

i.e. F= 10* 9.8 = 98 N

then X= F/K = 98/400 = 0.245 m

F/s = E ---> mg/h = E ---> 10 x 9,8 / s = 400 ----> s = 98/400 = 0.245. Answer : 2.45E-1

Force=-(force constant)(distance) Hooke's Law

Therefore, distance=-(10 kg x -9.8 ms-2)/ (400 kgms-2) =0.245 m

I must stretch the band so it applies a force equivalent to the gravitational force on a 10 kg weight is:

$\implies \frac{10 * 9.8}{400}$

$\implies \frac{98}{400} \implies \boxed{0.245 meters}$

Anyone watch that there is two "the" in the Middle of the Question?

F=kx=W=mg W=10[9.8]=400x Hence,x=0.245m