Question

When an archer pulls an arrow back in his bow, he is storing potential energy in the stretched bow. Compute the potential energy stored in the bow, if the arrow of mass 4.70 10-2 kg leaves the bow with a speed of 48.0 m/s. Assume that mechanical energy is conserved.

Answer 1

The potential energy stored in the bow when an arrow of mass 4.70 × 10^-2 kg is shot with a speed of 48.0 m/s is found to be 54.144 joules, assuming mechanical energy is conserved.

Step-by-step explanation:

When an archer pulls an arrow back in a bow, potential energy is stored in the stretched bow. To compute the stored potential energy, we assume that mechanical energy is conserved. Given that the mass of the arrow is 4.70 × 10-2 kg and it leaves the bow with a speed of 48.0 m/s, we can use the kinetic energy formula:

KE = ½ × m × v2

KE is the kinetic energy, m is the mass of the arrow, and v is the speed of the arrow. Plugging in the values:

KE = 0.5 × 4.70 × 10-2 kg × (48.0 m/s)2

KE = 0.5 × 4.70 × 10-2 kg × 2304 m2/s2

KE = 54.144 J

Therefore, the potential energy stored in the bow is 54.144 joules. This is the energy required to shoot the arrow with the given speed.