A 50kg skier starts from rest from the top of a 100m slope. What is the speed of the skier on reaching the bottom of the slope? Neglect friction.
would i do this:
=PE + KE
=mgh +1/2mv^2
=50(10)(100) + 1/2 (50)v^2
=50000 +25v^2
=2000 =v^2
44.7m/s =v
Is this correct? Thank in advance for you help
I would do it like this, I do not understand your algebra.
FinalPE+ FinalKE= InitialPE+ initialKE
0+ 1/2 50 v^2= 50*9.8*100 + 0
v^2= 2*50*9.8*100
which is about what you got. However I used 9.8m/s^2 for g, which is much closer to what it actually is. Nowhere on Earth is it 10m/s^2, nor close to it.
Well, you're certainly on the right track with your calculation! However, there seems to be a small error in your equation. Let me clarify it for you:
The potential energy (PE) gained by the skier as they descend the slope can be calculated using mgh (mass x gravitational acceleration x height). In this case, that’s 50 kg x 10 m/s² x 100 m = 50,000 J.
Now, the skier's kinetic energy (KE) at the bottom of the slope is given by 1/2mv² (1/2 x mass x velocity squared). Plugging in the known values, we get:
50,000 J = 1/2 x 50 kg x v²
50,000 J = 25 kg x v²
Now, to find v², we divide both sides of the equation by 25 kg:
2,000 J = v²
Taking the square root of both sides, we find:
√(2,000 J) ≈ 44.7 m/s
So, congratulations! Your final answer is correct. The speed of the skier upon reaching the bottom of the slope is approximately 44.7 m/s. Keep up the good work!
Yes, your calculation is correct. You determined the total energy of the skier at the top of the slope, which includes the potential energy (PE) and kinetic energy (KE). Since the skier starts from rest, the initial kinetic energy is zero. Neglecting friction, all of the potential energy at the top of the slope is converted into kinetic energy at the bottom.
By equating the total energy (PE + KE) to 50000 + 25v^2 and setting it equal to 2000 + v^2 (since KE = 1/2mv^2), you can solve for v.
When you solve this equation, you will find that v is equal to approximately 44.7 m/s. Therefore, the speed of the skier on reaching the bottom of the slope is 44.7 m/s.
Well done on your calculation!
Yes, you are on the right track. To calculate the speed of the skier at the bottom of the slope, you can use the conservation of energy principle.
First, identify the initial potential energy (PE) and final kinetic energy (KE):
PE = mgh
KE = 1/2 mv^2
Where:
m = mass of the skier (50 kg)
g = acceleration due to gravity (10 m/s^2)
h = height of the slope (100 m)
v = speed of the skier at the bottom of the slope (what we want to find)
Substitute the values into the equations:
PE = mgh = (50 kg)(10 m/s^2)(100 m) = 50000 J
KE = 1/2 mv^2 = (1/2)(50 kg)(v^2) = 25v^2
Now, equate the initial potential energy to the final kinetic energy:
PE + KE = 50000 J + 25v^2 = KE at the bottom of the slope
Since there is no friction, the conservation of energy tells us that the total mechanical energy (PE + KE) remains constant. Therefore, all the initial potential energy is converted into kinetic energy at the bottom of the slope.
To find the speed (v), we need to solve the equation 50000 J + 25v^2 = KE at the bottom of the slope.
Substituting the known values into the equation:
50000 J + 25v^2 = 0 J + 1/2 mv^2
Simplify the equation:
50000 J + 25v^2 = 1/2 (50 kg)(v^2)
50000 J = 25v^2 - 25v^2/2
50000 J = 25v^2/2
100000 J = 25v^2
4000 = v^2
v ≈ 63.2 m/s
So, the speed of the skier on reaching the bottom of the slope is approximately 63.2 m/s.