Question

A spring of negligible mass has force constant of 1600 Newtons per meter. (a) How far must the spring be compressed for 3.20 Joules of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a 1.20 kilogram book onto it from a height of 0.80 meters above the top of the spring. Find the maximum distance the spring will be compressed.

Answer 1

Answer:

a) x=63.0 $x10^{-3} m$ or 6.3cm

b) x=116.0 $x10^{-3} m$ or 11.6cm

Explanation:

a).

The elastic potential energy is modeling by equation :

$U1=\frac{1}{2}*k*x^{2} \\K=1600 \frac{N}{m}\\ U=3.2J\\m=1.2kg\\x^{2}=\frac{2*U}{k}\\x=\sqrt{\frac{2*3.2J}{1600 \frac{N}{m}}} \\x=\sqrt{4x10^{-3} m^{2}}\\ x=0.06324m$

b).

The work energy theorem explain which work is done in this case. the motion began from the rest so K1=K2 equal zero, Ug1 is no yet done and U2is also zero because is the potential energy

$Ug=K2\\mgy=\frac{1}{2}*k*x^{2} \\but \\y=h+x=0.80m+x\\m*g*(0.80m+x)=\frac{1}{2}*k*x^{2}$

Solving for x

$2*(m*g(h+x))=k*x^{2} \\k*x^{2}-(2*m*g*x)-(2*m*g*h)=0\\1600x^{2} -2*1.2kg*9.8\frac{m}{s^{2}}*x-2*1.2kg*0.80m=0\\1600x^{2} -23.52x-18.816m=0$

$x=-\frac{b+/-\sqrt{b^{2}-4*a*c } }{2*a} \\x=-\frac{-23.52+/-\sqrt{-23.52^{2}-4*1600*-18.816 } }{2*a} \\x=11.79 +/- 173.9\\x1=-0.10 m\\x2=0.116$

The negative is discard so

x=0.116m