Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula [tex]K E=\frac{1}{2} m v^2[/tex], where [tex]m[/tex] is the mass and [tex]v[/tex] is the speed (velocity). Assume that the speed of the soda bottle falling from a height of 0.8 m will be [tex]4 m / s[/tex], and use this speed for each calculation. Record your calculations in Table A of your Student Guide.
When the mass of the bottle is 0.125 kg, the KE is [tex]kg \cdot m ^2 / s ^2[/tex].
When the mass of the bottle is 0.250 kg, the KE is [tex]kg \cdot m ^2 / s ^2[/tex].
When the mass of the bottle is 0.375 kg, the KE is [tex]kg \cdot m ^2 / s ^2[/tex].
When the mass of the bottle is 0.500 kg, the KE is [tex]kg \cdot m ^2 / s ^2[/tex].
Answer (1)
Calculate the kinetic energy for m = 0.125 kg: K E 1 = 2 1 ( 0.125 ) ( 4 2 ) = 1 k g ⋅ m 2 / s 2 .
Calculate the kinetic energy for m = 0.250 kg: K E 2 = 2 1 ( 0.250 ) ( 4 2 ) = 2 k g ⋅ m 2 / s 2 .
Calculate the kinetic energy for m = 0.375 kg: K E 3 = 2 1 ( 0.375 ) ( 4 2 ) = 3 k g ⋅ m 2 / s 2 .
Calculate the kinetic energy for m = 0.500 kg: K E 4 = 2 1 ( 0.500 ) ( 4 2 ) = 4 k g ⋅ m 2 / s 2 . The kinetic energies are 1 , 2 , 3 , 4 k g ⋅ m 2 / s 2 .
Explanation
Understanding the Problem We are asked to calculate the kinetic energy (KE) of a bottle with different masses, using the formula K E = f r a c 1 2 m v 2 , where m is the mass and v is the speed. We are given that the speed v is constant at 4 m / s for all calculations. We need to calculate the KE for four different masses: 0.125 kg, 0.250 kg, 0.375 kg, and 0.500 kg.
Calculating KE for m = 0.125 kg First, let's calculate the kinetic energy when the mass is 0.125 kg. Using the formula K E = f r a c 1 2 m v 2 , we have: K E 1 = 2 1 ( 0.125 k g ) ( 4 m / s ) 2 = 2 1 ( 0.125 ) ( 16 ) = 0.125 ∗ 8 = 1 k g ⋅ m 2 / s 2
Calculating KE for m = 0.250 kg Next, let's calculate the kinetic energy when the mass is 0.250 kg: K E 2 = 2 1 ( 0.250 k g ) ( 4 m / s ) 2 = 2 1 ( 0.250 ) ( 16 ) = 0.250 ∗ 8 = 2 k g ⋅ m 2 / s 2
Calculating KE for m = 0.375 kg Now, let's calculate the kinetic energy when the mass is 0.375 kg: K E 3 = 2 1 ( 0.375 k g ) ( 4 m / s ) 2 = 2 1 ( 0.375 ) ( 16 ) = 0.375 ∗ 8 = 3 k g ⋅ m 2 / s 2
Calculating KE for m = 0.500 kg Finally, let's calculate the kinetic energy when the mass is 0.500 kg: K E 4 = 2 1 ( 0.500 k g ) ( 4 m / s ) 2 = 2 1 ( 0.500 ) ( 16 ) = 0.500 ∗ 8 = 4 k g ⋅ m 2 / s 2
Final Answer Therefore, the kinetic energies for the given masses are: When the mass of the bottle is 0.125 kg, the KE is 1 k g ⋅ m 2 / s 2 .
When the mass of the bottle is 0.250 kg, the KE is 2 k g ⋅ m 2 / s 2 .
When the mass of the bottle is 0.375 kg, the KE is 3 k g ⋅ m 2 / s 2 .
When the mass of the bottle is 0.500 kg, the KE is 4 k g ⋅ m 2 / s 2 .
Examples
Understanding kinetic energy is crucial in many real-world scenarios. For instance, when designing vehicles, engineers must consider the kinetic energy involved in collisions to develop effective safety measures. Similarly, in sports, understanding kinetic energy helps athletes optimize their performance, such as maximizing the energy transferred to a ball during a swing or kick. Moreover, kinetic energy principles are applied in renewable energy technologies, like wind turbines, to convert the kinetic energy of wind into electricity. These examples demonstrate the practical significance of kinetic energy in engineering, sports, and sustainable energy solutions.
