What is the kinetic energy (KE) of a 0.155 kg arrow shot upward at 31.4 m/s when it is 30.0 m above the ground? Thank you in advance for your help!
Physics · Middle School · Thu Feb 04 2021
Answered on
To calculate the kinetic energy (KE) of the arrow, we use the formula for kinetic energy:
\[ KE = \frac{1}{2}mv^2 \]
where: \( m \) is the mass of the arrow (in kilograms), \( v \) is the velocity of the arrow (in meters per second).
Substituting the given values:
\( m = 0.155 \) kg (mass of the arrow), \( v = 31.4 \) m/s (velocity of the arrow when shot upward).
\[ KE = \frac{1}{2} \times 0.155 \, \text{kg} \times (31.4 \, \text{m/s})^2 \] \[ KE = \frac{1}{2} \times 0.155 \times 986.36 \, \text{m}^2/\text{s}^2 \] \[ KE = 0.0775 \times 986.36 \] \[ KE = 76.5431 \, \text{J} \]
Therefore, the kinetic energy of the arrow when it is 30.0 m above the ground is approximately 76.54 joules.