Problem 1: Gravitational Potential Energy A 5 kg object is lifted to a height of 10 meters. Calculate the gravitational potential energy of the object.Problem 2: Kinetic EnergyA car with a mass of 1500 kg moves at a speed of 20 m/s. Calculate its kinetic energy.Problem 3: Converting Potential Energy to Kinetic EnergyA 2 kg ball is dropped from a height of 15 meters. Assuming no air resistance, calculate the speed of the ball just before it hits the ground.Solution:1. Calculate the Potential Energy at the top:1. Since there is no air resistance, all the potential energy converts to kinetic energy:Problem 4: Height from Potential EnergyA 10 kg object has a potential energy of 500 J. Calculate the height at which the object is located.Problem 5: Kinetic Energy and SpeedAn object has a kinetic energy of 800 J and a mass of 4 kg. Calculate its speed.
Answer (1)
1. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & m = \text{5 kg} \\ & h = \text{10 m} \end{aligned}[/tex]Step 2: Calculate the gravitational potential energy.[tex]\begin{aligned} GPE & = mgh \\ & = (\text{5 kg})(\text{9.8 m/s}^2)(\text{10 m}) \\ & = \boxed{\text{490 J}} \end{aligned}[/tex]Hence, the gravitational potential energy of the object is 490 J.2. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & m = \text{1,500 kg} \\ & v = \text{20 m/s} \end{aligned}[/tex]Step 2: Calculate the kinetic energy.[tex]\begin{aligned} KE & = \frac{1}{2}mv^2 \\ & = \frac{1}{2}(\text{1,500 kg})(\text{20 m/s})^2 \\ & = \boxed{\text{300,000 J}} \end{aligned}[/tex]Hence, the kinetic energy is 300,000 J.3. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & m = \text{2 kg} \\ & h = \text{15 m} \end{aligned}[/tex]Step 2: Find the potential energy.[tex]\begin{aligned} PE & = mgh \\ & = (\text{2 kg})(\text{9.8 m/s}^2)(\text{15 m}) \\ & = \text{294 J} \end{aligned}[/tex]Step 3: Determine the kinetic energy.KE = PE (Assuming no air resistance)KE = 294 JStep 4: Calculate the speed.[tex]\begin{aligned} KE & = \frac{1}{2}mv^2 \\ 2KE & = mv^2 \\ mv^2 & = 2KE \\ \frac{mv^2}{m} & = \frac{2KE}{m} \\ v^2 & = \frac{2KE}{m} \\ \sqrt{v^2} & = \sqrt{\frac{2KE}{m}} \\ v & = \sqrt{\frac{2KE}{m}} \\ & = \sqrt{\frac{2(\text{294 J})}{\text{2 kg}}} \\ & = \boxed{\text{17.15 m/s}} \end{aligned}[/tex]Hence, the speed is 17.15 m/s.4. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & PE = \text{500 J} \\ & m = \text{10 kg} \end{aligned}[/tex]Step 2: Calculate the height.[tex]\begin{aligned} PE & = mgh \\ mgh & = PE \\ \frac{mgh}{mg} & = \frac{PE}{mg} \\ h & = \frac{PE}{mg} \\ & = \frac{\text{500 J}}{(\text{10 kg})(\text{9.8 m/s}^2)} \\ & = \boxed{\text{5.10 m}} \end{aligned}[/tex]Hence, the height of the object is 5.10 m as located.5. SOLUTION:Step 1: List the given values.[tex]\begin{aligned} & m = \text{4 kg} \\ & KE = \text{800 J} \end{aligned}[/tex]Step 2: Calculate the speed.[tex]\begin{aligned} KE & = \frac{1}{2}mv^2 \\ 2KE & = mv^2 \\ mv^2 & = 2KE \\ \frac{mv^2}{m} & = \frac{2KE}{m} \\ v^2 & = \frac{2KE}{m} \\ \sqrt{v^2} & = \sqrt{\frac{2KE}{m}} \\ v & = \sqrt{\frac{2KE}{m}} \\ & = \sqrt{\frac{2(\text{800 J})}{\text{4 kg}}} \\ & = \boxed{\text{20 m/s}} \end{aligned}[/tex]Hence, the speed is 20 m/s.
