A 0.05 kg ball is attached to the end of a 0.5 m long string. The ball is spun around a circle with a period of 0.20 s. What is the tension force in the string?

The ball spins once around the circle in 0.20 s, meaning it travels a distance equal to its circumference in that time, giving it a linear speed of

v = (2π (0.5 m)) / (0.20 s) = 5π m/s ≈ 15.708 m/s

Use this compute the magnitude of the centripetal acceleration a :

a = v²/ (0.5 m) = 50π² m/s² ≈ 493.48 m/s²

Use Newton's second law to compute the mangitude of the tension F in the string:

F = (0.05 kg) a = 5/2 π² N ≈ 24.674 N ≈ 25 N

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

i don't understand. explain it clearly