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Question Video: Finding the Speed of Body Sliding on a Smooth Inclined Plane at the Bottom Mathematics

A body started to slide down a smooth inclined plane of height 504 cm from its top. Find its speed when it reached the bottom. Take 𝑔 = 9.8 m/sΒ².


Video Transcript

A body started to slide down a smooth inclined plane of height 504 centimeters from its top. Find its speed when it reaches the bottom. Take 𝑔 equals 9.8 meters per second squared.

We can call the height difference that the body moves through, 504 centimeters, β„Ž. Knowing that the acceleration due to gravity is 9.8 meters per second squared, we want to solve for the speed of the body when it reaches the bottom of this height difference. We’ll call that speed 𝑣.

We can start on our solution by drawing a diagram of the body’s motion. A body slides down a vertical distance β„Ž, which we’re told is 504 centimeters. It slides on a smooth surface. So no energy is lost due to friction. We want to solve for the body’s speed 𝑣 after it’s descended that distance. And to do that, we can recall the principle of energy conservation.

This principle tells us that the initial energy in a system, 𝐸 sub 𝑖, is equal to the final energy in that system. We can expand on this by saying that the sum of the potential and kinetic energy a system has initially is equal to the sum of those two energy types that it has at the end.

In our example, the initial moment is when our body is at the top of the height and not yet in motion. And our final state is when the body is all the way at the bottom of the height and moving with a speed 𝑣. When we write out this energy balance equation for our scenario, we can simplify it because some of our terms are zero.

Looking at our initial state, at first, our body is not in motion. So its speed is zero. And therefore, its kinetic energy is zero. And at the end of our descent, when our body is all the way at the bottom of the height β„Ž, its potential energy is equal to zero.

Our equation reduces to the initial potential energy of our body is equal to its final kinetic energy. If we recall that gravitational potential energy is given mathematically as π‘š times 𝑔 times β„Ž and that kinetic energy is given as one-half π‘šπ‘£ squared, so we can write that π‘šπ‘”β„Ž, the initial potential energy of our body, is equal to the final kinetic energy, one-half π‘šπ‘£ squared.

Since the mass of the body is common to both sides, it cancels out. And when we rearrange this equation to solve for the speed 𝑣, we find it’s equal to the square root of two times 𝑔 times β„Ž.

Both 𝑔, the acceleration due to gravity, and β„Ž, the height of our elevation, are given. So we’re ready to plug in and solve for 𝑣. When we do plug in, we’re careful to convert our height β„Ž into units of meters.

If we then multiply both the numerator and denominator of this expression by 100 and then bring the factor of 100 in the numerator underneath the square root, it becomes a factor of 10000, which when we multiply through becomes on top the square root of 987840. If we factor a four outside of the square root in the numerator, then our denominator reduces to 25. And the expression under the square root becomes 61740.

This number is equal to 42 squared times 35. So we can bring a factor of 42 outside the square root sign. And this leaves us with a simplified expression for 𝑣 in units of meters per second. The speed of the body after it’s descended a height β„Ž is 42 times the square root of 35 divided by 25 meters per second.

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