A 0.4 kg mass connected to a low mass horizonal spring of stiffness 200 N/m slides on a very low friction surface. You observe the system at a moment when the spring is compressed 10 cm and the speed of the mass is 3 m/s. What will the kinetic energy of the mass be when the spring is stretched 5 cm?

The kinetic energy of the mass when the spring is stretched 5 cm is 2.55 J

Explanation:

Here we have

The mass on the spring = 0.4 kg

Spring stiffness = 200 N/m

Speed when compressed 10 cm = 3 m/s

E = 0.5·m·v² + 0.5·k·x² = 0.5·k·x²

Therefore

When x = 10 cm = 0.1 m, we have

E = 0.5·0.4·3² + 0.5·200·0.1² = 0.5·200·x² = 2.8

x² = 2.8/100  = 0.028,

x = 0.167 m

When the mass is at 5 cm we have

E = 0.5·0.4·v² + 0.5·200·0.05² = 2.8 J

Therefore, the kinetic energy = 0.5·0.4·v² = 2.8 - 0.25 = 2.55 J

The kinetic energy of the mass when the spring is stretched 5 cm = 2.55 J.

answer: question1

1,000 tons

question2

3: 8

question 3

store d

step-by-step explanation:

f=ma newton's second law 7x8=56n