Atoy of mass 0.170-kg is undergoing shm on the end of a horizontal spring with force constant k = 250 n/m . when the toy is a distance 0.0110 m from its equilibrium position, it is observed to have a speed of 0.400 m/s .
a)what is the toy's total energy at any point of its motion? express your answer with the appropriate units.
b)what is the toy's amplitude of the motion? express your answer with the appropriate units.
c)what is the toy's maximum speed during its motion? express your answer with the appropriate units.

a) Em = 2.87 10⁻² J , b) A = 0.0152 m , c) v = 0.581 m / s

Explanation:

a) For this problem we will use the energy ratio. Mechanical energy is conserved or we can calculate it in one place. We fear energy for all places. Caicule the mechanical energy in the place where they give data

    Em = K +

    Em = ½ m v² + ½ k x²

    Em = ½ 0.170 0.400² + ½ 250 0.0110²

    Em = 0.0136 + 0.0151

    Em = 0.0287 J

    Em = 2.87 10⁻² J

b) to calculate the amplitude we use the energy

    Em = ½ k A²

    A = √ 2 Em / k

    A = √(2 0.0287 / 250)

    A = 0.01515 m

    A = 0.0152 m

c) the maximum speed

At the point where the elongation is zero, midpoint of the movement, all energy is kinetic and this is the maximum value of the velocity

    Em = ½ m v²

    v = √2 Em / m

    v = √(2 0.0287 / 0.170)

    v = 0.581 m / s

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer