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Question Video: Finding the Change in Gravitational Potential Energy Mathematics

A particle of mass 281 g was projected at 37 cm/s up the line of greatest slope of a smooth plane inclined to the horizontal at an angle whose sine is 10/11. Determine the change in the particleโ€™s gravitational potential energy from the moment it was projected until its speed became 29 cm/s.

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Video Transcript

A particle of mass 281 grams was projected at 37 centimeters per second up the line of greatest slope of a smooth plane inclined to the horizontal at an angle whose sine is ten elevenths. Determine the change in the particleโ€™s gravitational potential energy from the moment it was projected until its speed became 29 centimeters per second.

We will begin by sketching a diagram to model the scenario. We are told that a particle of mass 281 grams was projected at 37 centimeters per second up the plane. At some point on the journey, the speed of the particle is 29 centimeters per second. We are asked to calculate the change in gravitational potential energy during this motion. We are also told that the sine of the angle of inclination ๐›ผ is equal to ten elevenths.

We will let the initial potential energy ๐‘ƒ sub i be equal to zero. And we therefore just need to calculate the final potential energy ๐‘ƒ sub f, which occurs when the speed of the particle is 29 centimeters per second. Since the slope is smooth, no energy is dissipated. And from the conservation of energy, we know that the increase in potential energy is equal to the decrease in kinetic energy. We can therefore conclude that the change in kinetic energy plus the change in potential energy equals zero, where ฮ”KE and ฮ”PE denote the change in kinetic and potential energy, respectively.

We know that the kinetic energy of an object is equal to a half ๐‘š๐‘ฃ squared, where ๐‘š is the mass and ๐‘ฃ the velocity. The change in kinetic energy is therefore equal to a half ๐‘š๐‘ฃ sub f squared minus a half ๐‘š๐‘ฃ sub i squared, where ๐‘ฃ sub f is the final velocity and ๐‘ฃ sub i the initial velocity. We can factor out a half ๐‘š, giving us a half ๐‘š multiplied by ๐‘ฃ sub f squared minus ๐‘ฃ sub i squared. And substituting the values from this question, we have the change in kinetic energy is equal to a half multiplied by 281 multiplied by 29 squared minus 37 squared. Typing this into our calculator gives us negative 74,184.

We can now calculate the change in potential energy by adding 74,184 to both sides of our equation. In this question, the mass was measured in grams and the velocity in centimeters per second. Whilst we couldโ€™ve converted these to the standard units of kilograms and meters per second so that the energy would be measured in joules, by using the mass in grams and velocity in centimeters per second, the potential energy will be measured in ergs. Since the increase in potential energy is equal to the decrease in kinetic energy, the change in the particleโ€™s potential energy from the moment it was projected until its speed became 29 centimeters per second is 74,184 ergs.

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