A skateboarder at a skate park rides along the path shown in (Figure 1) .
a=2.7 m and b=1.0m
If the speed of the skateboarder at point A is v=1.6m/s, what is her speed at point B? Assume that friction is negligible.
Solve the downhill speed first:
Vf = sqrt Vi^2+2*mh => Vf=sqrt 1.4m/s^2+2*(9.81)*(2.7m)=7.42m/s
Next solve uphill speed. Use Vf from downhill speed as Vi for calculating final speed going uphill.
Use the same formula as before but replace height (h) with the delta in height. We're calculating from the ground level up 1 meter so the initial height =0m, final height =1m so delta y = -1m. Makes sense this is a negative number since now the skateboarder is going uphill and losing speed.
Vf= sqrt 7.42m/s^2+2*(9.81)*(-1)= 5.95m/s
Mistake in that answer.
First equation should have read: Vf = sqrt Vi^2+2*gh. Sorry.
Well, it sounds like this skateboarder is really cruising along! But let's not get too ahead of ourselves, we need to calculate her speed at point B.
Using the law of conservation of energy, we can solve this riddle. At point A, the skateboarder has only gravitational potential energy. And at point B, she has both gravitational potential energy and kinetic energy.
Now, the gravitational potential energy at point A can be calculated using the formula:
PE = mgh
Where m is the mass of the skateboarder, g is the acceleration due to gravity, and h is the height difference between point A and the ground.
But since the question didn't provide any information about the height, let's say the skateboarder is on top of a high stack of pancakes. Because who doesn't love pancakes? So, let's say h = stack of pancakes.
Now, let's move on to the gravitational potential energy at point B. At point B, the skateboarder has lost some of her potential energy but gained kinetic energy. And the sum of these two energies remains constant.
Using this information, we can say:
PE_A + KE_A = PE_B + KE_B
But since the skateboarder's initial kinetic energy KE_A is 0, we can simplify it further:
PE_A = PE_B + KE_B
Now, we can convert the potential energy into kinetic energy at point B using the formula:
KE = (1/2)mv^2
So, we can rewrite our equation as:
mgh_A = (1/2)mv_B^2
Canceling out the mass on both sides of the equation, we get:
gh_A = (1/2)v_B^2
And finally, solving for v_B:
v_B = √(2gh_A)
But wait, we still need the height! Ah, let's make it easy and say the height is the same as the skateboarder's enthusiasm for landing a sick trick. So, h_A = enthusiasm.
Now, substituting this value in our equation, we get:
v_B = √(2g*enthusiasm)
And there you have it! The skateboarder's speed at point B is equal to the square root of twice her enthusiasm multiplied by the acceleration due to gravity.
Remember, safety first! Keep your helmet on and don't forget to have fun while skateboarding.
To find the speed of the skateboarder at point B, we can analyze the conservation of energy and the conservation of mechanical energy.
The conservation of energy states that the total mechanical energy of a system remains constant unless acted upon by external forces. In this case, since friction is negligible, we can assume that there are no external forces acting on the skateboarder.
At point A, the skateboarder has both kinetic energy (due to her speed) and gravitational potential energy (due to her height above the ground).
The kinetic energy (KE) at point A is given by the equation KE = (1/2)mv², where m is the mass of the skateboarder and v is her speed. However, since mass is not provided and not relevant to the problem, we can ignore it for now.
The gravitational potential energy (PE) at point A is given by the equation PE = mgh, where m is the mass of the skateboarder, g is the acceleration due to gravity, and h is the height of the skateboarder above the ground. In this case, h is given by a - b, which is 2.7 m - 1.0 m = 1.7 m.
Since the total mechanical energy is conserved, the sum of the kinetic and potential energies at point A is equal to the sum of the kinetic and potential energies at point B.
So, at point A, we have:
KE₁ + PE₁ = KE₂ + PE₂
Substituting in the given values:
(1/2)(v₁)² + mgh₁ = (1/2)(v₂)² + mgh₂
We are given the speed at point A (v₁ = 1.6 m/s) and the heights at points A and B (h₁ = 1.7 m, h₂ = 0 m since point B is at ground level).
The equation now becomes:
(1/2)(1.6)² + mgh₁ = (1/2)(v₂)² + mgh₂
Simplifying further:
(1/2)(2.56) + m(9.8)(1.7) = (1/2)(v₂)²
1.28 + 16.7m = (1/2)(v₂)²
Since we don't have the mass of the skateboarder, we can't solve for the exact value of the speed at point B. We can, however, find a relationship between the speeds at points A and B.
Assuming the mass remains constant, we can cancel out the mass term:
1.28 = (1/2)(v₂)²
Simplifying again:
2.56 = (v₂)²
Taking the square root of both sides:
√(2.56) = v₂
So, the speed of the skateboarder at point B is approximately 1.6 m/s.