Missy Diwater, the former platform diver for the Ringling Brother’s Circus had a kinetic energy of 15 000 J just prior to hitting the bucket of water. If Missy’s mass is 50 kg, then what is her speed? Solution: According to energy conservation, the kinetic energy at the bottom of the dive (15,000J) is equal to her gravitational potential energy before the dive. We can use this fact to find her dive height: PE = mgh h = PE/mg = 15,000J / (50kg)(9. 81m/s? ) 31m (rounded) Her speed can also be found from energy conseration: E(final) = E(initial) 0. 5mv? = mgh v = v[2gh] = v[2(9. 81m/s? )(31m) = 25m/s 2. A 750-kg compact car moving at 100 km/hr has approximately 290 000 Joules of kinetic energy. What is the kinetic energy of the same car if it is moving at 50 km/hr? Solution: KE =v^ 2 (Kinetic Energy = speed ^2 If the speed is reduce by a factor of 2 (as in form 100 km/hr) then the KE will reduce by a factor 4. Thus,the new KE = 290 000 J / 4 KE = 72 500 J 3.

A cart is loaded with a brick and pulled at constant speed along an inclined plane of an angle of 30o to the height of a seat-top. If the mass of the loaded cart is 3. 0 kg and the inclined distance of the seat top is 0. 45 meters, then what is the potential energy of the loaded cart at the height of the seat-top? Solution : PE = mgh PE = 3 kg x 10 m/s/s x 0. 45m PE = 13. 5 J 4. A 75kg trampoline artist jumps vertically downward from the top of a platform with a speed of 5m/s. How fast is he going as he lands on the trampoline 2m below?

### The Report on Gravitational Potential Energy and Kinetic Energy

Purpose: The purpose of this is to learn how to calculate gravitational potential energy and kinetic energy, and to see how gravitational potential energy and kinetic energy affect the car during the lab. Define: Gravitational Potential Energy- Energy an object possesses because of its position in a gravitational field Kinetic Energy- The energy possessed by a body because of its motion, equal to ...

If the trampoline behaves like a spring of spring constant 5. 2E104 N/m, how far does he depress it? Soluiton : a) s = 1/2(u+v)t 2. 0m = 0. 5 * 5m/s * t 2. 0m = 10 * t t = 2. 0m/20 t = 0. 1s b) Hooke’s Law states F=kx x is the displacement of the spring (depression) F = Restoring force k = spring constant Rearrange. x = F/k What is the force upon hitting the trampoline? We have the mass so let’s work out the acceleration. Acceleration = velocity/time Acceleration = 5/0. 1 = 50m/s^2 F=ma F = 75*50 = 3750N Substitute into Hooke’s Law x = 3750/(5. 2*10^4N-m) = 0. 072m of depression