if a bean of mass 2.0 g jumps 1.0 cm from your hand into the air, how much potential energy has it gained in reaching its highest point? What is its speed as the bean lands back in the palm of your hand?
PE gained=mass*g*heightinmeters mass is in kg g=9.8Nt/kg
find speed from
1/2 mass*v^2=mass*g*heightfell
use kg, meters, seconds as units.
To find the potential energy gained by the bean as it reaches its highest point, we can use the formula for gravitational potential energy:
Potential Energy = mass * gravity * height
Given:
mass (m) = 2.0 g = 0.002 kg (since 1 g = 0.001 kg)
gravity (g) = 9.8 m/s^2 (approximate acceleration due to gravity on Earth)
height (h) = 1.0 cm = 0.01 m (since 1 cm = 0.01 m)
Substituting the values into the formula:
Potential Energy = 0.002 kg * 9.8 m/s^2 * 0.01 m
Potential Energy = 0.000196 J (Joules)
Therefore, the bean has gained approximately 0.000196 Joules of potential energy in reaching its highest point.
To find the speed of the bean as it lands back in the palm of your hand, we need to consider the conservation of mechanical energy. At the highest point, all the potential energy gained is converted into kinetic energy.
Thus, potential energy at highest point = kinetic energy at landing point
Potential Energy = Kinetic Energy
mgh = (1/2)mv^2
Where:
m = mass of the bean (0.002 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height gained (0.01 m)
v = final velocity at landing point (what we want to find)
Rearranging the formula:
v^2 = 2gh
v = √(2gh)
Substituting the values:
v = √(2 * 9.8 m/s^2 * 0.01 m)
v ≈ √0.196
v ≈ 0.443 m/s
Therefore, the speed of the bean as it lands back in the palm of your hand is approximately 0.443 m/s.