Potential Energy Questions PDF

A tree branch is found to obey Hooke's Law. When a 20 kg load is hung from the end of it, the branch sags a distance of 10 cm. If, instead, a 40 kg load is hung from the same place, how much will the bag sag? F = kx k = F1/x = (20)(9.8) / 0.1 = 1960 N/m x = F2/k = (40)(9.8) / 1960 = 0.2 m = 20 cm If...

A tree branch is found to obey Hooke's Law. When a 20 kg load is hung from the end of it, the branch sags a distance of 10 cm. If, instead, a 40 kg load is hung from the same place, how much will the bag sag? F = kx k = F1/x = (20)(9.8) / 0.1 = 1960 N/m x = F2/k = (40)(9.8) / 1960 = 0.2 m = 20 cm If a force of 10 N stretches a certain spring by 4 cm, how much stretch will occur for an applied force of 15 N? F = kx k = F/x = 10/0.04 = 250 N/m x = F/k = 15/250 = 0.06 m = 6 cm A spring with a force constant of 5.2 N/m has a relaxed length of 2.45 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.57 m. Calculate the elastic potential energy stored in the spring. x = L - L0 = 3.57 - 2.45 = 1.12 m PE(spring) = ½(k)(x)^2 PE(spring) = ½(5.2)(1.12^2) PE(spring) ≈ 3.3 J The staples inside a stapler are kept in place by a spring with a relaxed length of 0.115 m. If the spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its length is 0.150 m? PE(spring) = ½(k)(x)^2 x = L - L0 = 0.150 - 0.115 = 0.035 m PE(spring) = ½(51.0)(0.035^2) PE(spring) ≈ 0.031 J A 40.0 kg child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions: 1. when the ropes are horizontal. PE(grav) = mgh PE(grav) = (40.0)(9.8)(2.00) = 784 J 2. when the ropes make a 30.0° angle with the vertical. Hint: use the height above the child's lowest position to calculate. h = r(1 - cos θ) h = 2.00(1 - cos 30.0) h = 2.00(1 - 0.866) = 0.268 m PE(grav) = mgh = (40.0)(9.8)(0.268) ≈ 105 J 3. at the bottom of the circular arc. PE(grav) = mgh = (40.0)(9.8)(0) = 0.00 J