Question

A horizontal spring with spring constant of 9.80 N/m is attached to a block with a mass of 1.20 kg that sits on a frictionless surface. When the block is 0.345 m from its equilibrium position, it has a speed of 0.540 m/s.(a) What is the maximum displacement of the block from the equilibrium position?m(b) What is the maximum speed of the block?m/s(c) When the block is 0.200 m from the equilibrium position, what is its speed?m/s

Answer 1

Answer:

Explanation:

Given

at certain point displacement and velocity is 0.345 m and 0.54 m/s

Therefore Potential and Kinetic Energy associated is

$U=\frac{1}{2}kx^2$

$U=\frac{1}{2}\times 9.8\times (0.345)^2$

$U=0.583 J$

Kinetic Energy $K=\frac{1}{2}mv^2$

$k=\frac{1}{2}\times 1.2\times (0.54)^2=0.174 J$

Total Energy $T=U+K=0.583+0.174=0.757 J$

Total Energy is conserved hence at maximum displacement  all energy will be Potential energy

$T=\frac{1}{2}kA^2$

where A=maximum displacement

$0.757=\frac{1}{2}\times 9.8\times A^2$

$0.1544=A^2$

$A=0.393 m$

Maximum speed occurs at equilibrium Position where Potential Energy is zero

thus $T=\frac{1}{2}mv^2$

$0.757=\frac{1}{2}\times 1.2\times v_{max}^2$

$v_{max}=1.123 m/s$

(c)When block is at 0.2 m from Equilibrium speed then its Potential Energy is

$U=\frac{1}{2}kx^2=\frac{1}{2}\times 9.8\times (0.2)^2=0.196 J$

$T=U+K$

$K=0.757-0.196=0.561 J$

$K=\frac{1}{2}mv^2$

$0.561=\frac{1}{2}\times 1.2\times v^2$

$v=0.966 m/s$