Question Video: Finding the Kinetic Energy of a Body at the Bottom of a Smooth Inclined Plane Mathematics

A particle of mass 225 g started moving from the top of a smooth plane inclined to the horizontal at an angle whose sine is 1/3. Given that the plane was 48 m long, find the kinetic energy of the particle when it reached the bottom of the plane. Take 𝑔 = 9.8 m/sΒ².

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Video Transcript

A particle of mass 225 grams started moving from the top of a smooth plane inclined to the horizontal at an angle whose sine is one-third. Given that the plane was 48 meters long, find the kinetic energy of the particle when it reached the bottom of the plane. Take 𝑔 equal to 9.8 meters per second squared.

We know that the kinetic energy of a particle is equal to a half π‘šπ‘£ squared, where π‘š is the mass of the particle in kilograms and 𝑣 is the velocity in meters per second. The kinetic energy will be measured in joules. We know that there are 1000 grams in a kilogram. Therefore, the mass of the particle is 0.225 kilograms as 225 divided by 1000 is 0.225. If we sketch a diagram showing the particle at the top of the plane, we know that the force acting vertically downward is equal to 0.225𝑔, where 𝑔 is the acceleration due to gravity equal to 9.8 meters per second squared. We know that the sin of angle 𝛼 is equal to one-third. We are also told that the plane is 48 meters long.

We know that the force driving the object down the plane will be equal to the component of the weight acting parallel to the plane. By creating a right-angled triangle, we can use our knowledge of trigonometry to calculate this force 𝐹. We know that the sine of any angle is equal to the opposite over the hypotenuse. Therefore, the sin of angle 𝛼 is equal to π‘₯ over 0.225𝑔. Multiplying both sides of this equation by 0.225𝑔 gives us π‘₯ is equal to 0.225𝑔 multiplied by one-third, the value of sin 𝛼. Typing this into the calculator, we get that π‘₯ is equal to 0.735. The force acting parallel to the plane is equal to 0.735 newtons.

We can now use Newton’s second law to calculate the acceleration of the particle. This states that the net sum of the forces is equal to the mass multiplied by the acceleration. 0.735 is equal to 0.225 multiplied by π‘Ž. Dividing both sides by 0.225, we get a value of π‘Ž equal to 49 over 15. The acceleration of the particle is 49 over 15 meters per second squared.

Our next step is to calculate the velocity of the particle when it reaches the bottom of the plane. In order to do this, we will use our equations of motion or SUVAT equations. We know that the displacement or distance the particle travels is 48 meters, the initial speed is zero meters per second, and the acceleration is equal to 49 over 15 meters per second squared.

We will use the equation 𝑣 squared is equal to 𝑒 squared plus two π‘Žπ‘ . Substituting in our values, we have 𝑣 squared is equal to zero squared plus two multiplied by 49 over 15 multiplied by 48. 𝑣 squared is therefore equal to 313.6. We can calculate the velocity of the particle by square rooting both sides of this equation. This gives us a value of 𝑣 equal to 17.708. This is the velocity of the particle when it reaches the bottom of the plane.

In this question, it will be more useful to use the fact that 𝑣 squared is equal to 313.6. We can substitute this into our kinetic energy formula, giving us one-half multiplied by 0.225 multiplied by 313.6. This gives us an answer of 35.28.

The kinetic energy of the particle when it reaches the bottom of the plane is 35.28 joules.

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