A 63 kg skier encounters a dip in the snow's surface that has a circular cross section with a radius of curvature of 12 m. If the skier's speed at point A in Figure 8-25 is 7.9 m/s, what is the normal force exerted by the snow on the skier at point B? Ignore frictional forces.
radius=12
h of dip=1.75m
We don't have access to your Figure 8-25. The normal force extered by the snow on the skier must equal the centripetal force MV^2/R , plus the component of the weight (M g) in the normal direction. Use conservation of energy to get V at point B.
R = 12 m
Why did the skier bring a ladder to the snowy mountain? Because they wanted to take their skiing to a whole new level! Now, let me calculate the normal force for you.
To find the normal force exerted by the snow on the skier at point B, we can use the principle of conservation of energy. Here are the steps to calculate it:
1. Determine the gravitational potential energy at point A:
The gravitational potential energy (PE) at any point in the dip is given by the equation:
PE = m * g * h
where m is the mass of the skier (63 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the dip (1.75 m).
So, PE at point A = 63 kg * 9.8 m/s^2 * 1.75 m = 1078.35 J
Note: At point A, the skier has all potential energy and no kinetic energy.
2. Determine the total mechanical energy at point A:
The total mechanical energy (E) at any point in the dip is the sum of its kinetic energy (KE) and potential energy (PE).
At point A, the skier's speed is given as 7.9 m/s. Therefore, the kinetic energy (KE) at point A is:
KE = 0.5 * m * v^2
where m is the mass of the skier (63 kg) and v is the speed (7.9 m/s).
So, KE at point A = 0.5 * 63 kg * (7.9 m/s)^2 = 1994.13 J
Total mechanical energy at point A (E) = KE + PE = 1994.13 J + 1078.35 J = 3072.48 J
3. Determine the total mechanical energy at point B:
Since there is no mention of frictional forces, we can assume that the mechanical energy is conserved throughout the dip, and hence, the total mechanical energy at point B will be the same as at point A (E = 3072.48 J).
4. Determine the gravitational potential energy at point B:
At point B, the skier is higher than at point A by the height of the dip (1.75 m).
PE at point B = m * g * h
where m is the mass of the skier (63 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the dip (1.75 m).
So, PE at point B = 63 kg * 9.8 m/s^2 * 1.75 m = 1078.35 J
5. Determine the normal force at point B:
The normal force (N) at any point is equal to the weight (mg) minus the gravitational force (mgcosθ). In this case, since there is no frictional force, the angle between the normal force and weight (θ) is 0 degrees.
Therefore, the normal force (N) at point B = weight (mg) - gravitational force (mgcosθ) = mg - mg*cos0 = mg
So, the normal force at point B = 63 kg * 9.8 m/s^2 = 617.4 N
Therefore, the normal force exerted by the snow on the skier at point B is 617.4 N.
To find the normal force exerted by the snow on the skier at point B, we need to consider the forces acting on the skier.
First, we'll calculate the acceleration of the skier at point B using the conservation of energy principle. The potential energy at point A is converted into kinetic energy at point B. Since we are ignoring frictional forces and assuming all the energy is conserved, we can write:
Potential energy at A = Kinetic energy at B
mgh = (1/2)mv^2
Where:
m = mass of the skier (63 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the dip (1.75 m)
v = speed at point A (7.9 m/s)
Substituting the given values, we have:
(63 kg)(9.8 m/s^2)(1.75 m) = (1/2)(63 kg)(v^2)
Using the units correctly, the equation becomes:
(63 kg)(9.8 m/s^2)(1.75 m) = (1/2)(63 kg)(7.9 m/s)^2
Next, we can solve for the speed at point B, v:
v^2 = [(63 kg)(9.8 m/s^2)(1.75 m)] / [(1/2)(63 kg)]
v^2 = (1225 N) / (31.5 kg)
v^2 ≈ 38.89 m^2/s^2
v ≈ √(38.89 m^2/s^2)
v ≈ 6.23 m/s
Now that we have the speed at point B (approximately 6.23 m/s), we can calculate the centripetal acceleration of the skier at point B using the formula:
a = v^2 / r
Where:
v = speed at point B (6.23 m/s)
r = radius of curvature (12 m)
Substituting the given values, we have:
a = (6.23 m/s)^2 / 12 m
a ≈ 3.23 m/s^2
Finally, the normal force exerted by the snow on the skier is equal to the skier's weight plus the centripetal force acting towards the center of the circular dip:
Normal force = weight + centripetal force
Normal force = (mass of skier)(acceleration due to gravity) + (mass of skier)(centripetal acceleration)
Normal force = (63 kg)(9.8 m/s^2) + (63 kg)(3.23 m/s^2)
Evaluating the expression, we get:
Normal force ≈ 617.4 N
Therefore, the normal force exerted by the snow on the skier at point B is approximately 617.4 N.