A 3 kg object falls from a height of 10 m. What is its velocity just before it hits the ground? How would this problem differ if the object falling from a height of 10 m only had a mass "m"?
1/2 mv^2=mgh
solve for velocity v
To find the velocity of the object just before it hits the ground, we can use the principles of energy conservation.
Step 1: Determine the potential energy.
The potential energy of an object at a height h is given by the formula: PE = m * g * h, where m is the mass of the object, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
For the given situation, the potential energy is: PE = 3 kg * 9.8 m/s² * 10 m = 294 J.
Step 2: Use conservation of energy.
The potential energy of the object is converted into kinetic energy as it falls. The kinetic energy is given by the formula: KE = (1/2) * m * v², where m is the mass of the object and v is its velocity.
Since energy is conserved, we can equate the potential energy to the kinetic energy, resulting in the equation: PE = KE.
Substituting the values, we have: 294 J = (1/2) * 3 kg * v².
Step 3: Solve for velocity.
Rearranging the equation, we find: v² = (2 * 294 J) / 3 kg.
Calculating this, we get: v² ≈ 196 m²/s².
Finally, taking the square root of both sides, we find that the magnitude of the velocity just before the object hits the ground is approximately: v ≈ 14 m/s.
Now, let's consider how the problem would differ if the mass of the object falling from a height of 10 m were denoted as "m" instead of having a specific value.
In this case, we would express the potential energy as: PE = m * g * h, where m is the mass (given as "m"), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height.
The expression for potential energy remains the same as before. The only difference is that, now, we can't determine the exact value of "m" since it is not provided.
Subsequently, when solving for the velocity using the conservation of energy equation, we would have an equation in terms of "m" and "v," without being able to calculate an actual numerical value for velocity.