Question

JA spring is used to launch na ball vertically into the air. The spring has a spring constant of 200 N/m and is compressed by 5cm. A ball of mass 10 g is placed just above the spring. Calculate: ) a)the energy stored in spring b)assumingthespringtransferallofitsenergytotheballthevelocityoftheballjustasitlaunches c)theheightreachedtheballassumingallthe EK isconvertedinto gpe

​

Answer 1

• 0.25 J
• √50 ≈ 7.07 m/s
• 2.55 m

Step-by-step explanation:

You want the energy stored in a spring with k = 200 N/m when it is compressed 5 cm, the velocity of a 10 g ball after that energy is transferred to it, and the height that ball reaches if it is launched upward.

a) Spring energy

The energy stored in a spring with constant k compressed by x is ...

PE = 1/2kx²

Using MKS units, we find the potential energy of the given spring to be ...

PE = 1/2(200 N/m)(0.05 m)² = 0.25 J

The energy stored in the spring is 0.25 joules.

b) Launch velocity

A 10 g ball with a kinetic energy of 0.25 J will have a velocity of ...

v = √(2KE/m)

v = √(2(0.25 kg·(m/s)²)/(0.01 kg)) = √50 m/s ≈ 7.07 m/s

The initial velocity of the ball will be √50 ≈ 7.07 m/s.

c) Maximum height

After the kinetic energy is converted to potential energy, the height will be ...

mgh = KE

h = KE/(mg) = (0.25 J)((0.01 kg)(9.8 m/s²) = (25/9.8) m ≈ 2.55 m

The ball will reach a height of about 2.55 meters.