A student has a weight of 650 N. While riding on a roller-coaster this same student has an apparent weight of 1.97 x 103 N at the bottom of a loop that has a radius of 17.0 m. The speed of the roller-coaster is
I just need the forumla, doesn't even have to be manipulated.
At the bottom of the loop, the apparent weight is
1970 = (Weight) + (Centripetal force)
= 650 + M*V^2/R
M = W/g = 66.33 kg
1320 = 66.33*V^2/17
Solve for V. It will be in m/s
Thanks so much!
Well, it seems like this student is having quite the thrilling ride on the roller-coaster! Now, let's talk about the formula you need. To determine the speed of the roller-coaster, we can make use of the apparent weight of the student at the bottom of the loop.
The formula you're looking for is given by:
apparent weight = actual weight + (mass x acceleration)
In this case, the apparent weight is 1.97 x 10^3 N, the actual weight is 650 N, and the acceleration is provided by the centripetal force required to keep the student moving in a circle of radius 17.0 m.
I hope you're ready for some calculations because we'll need to use a bit of math to determine the speed. Hang on tight, and let's do this!
The formula you can use to calculate the speed of the roller-coaster is the centripetal force formula:
F = (m * v^2) / r
where:
F is the centripetal force (apparent weight of the student),
m is the mass of the student,
v is the speed of the roller-coaster, and
r is the radius of the loop.
Please note that weight is the force of gravity acting on an object, and mass is a measure of the amount of matter in an object.
Is there anything else I can help you with?
The formula needed to solve this problem is the centripetal force formula:
F = m * a = m * v^2 / r
Where:
F is the centripetal force (in this case, the apparent weight of the student)
m is the mass of the student (which we will assume to be constant)
a is the centripetal acceleration
v is the velocity of the roller-coaster
r is the radius of the loop
In this case, the question provides us with the apparent weight of the student (F) at the bottom of the loop, the radius of the loop (r), and the weight of the student (m*g).
However, the formula given above is for the centripetal force, not the weight. To equate the two, we need to consider that the student's weight consists of the gravitational force (mg) acting vertically downward, and the normal force (N) provided by the roller coaster acting vertically upward.
So, by equating the net force at the bottom of the loop to the apparent weight, we have:
N - mg = F
Now, if we substitute F = m * v^2 / r, we get:
N - mg = m * v^2 / r
Now, if we simplify the equation, we can solve for v:
v^2 = g * r * (N/m - 1)
And finally:
v = sqrt(g * r * (N/m - 1))
Where:
v is the velocity of the roller-coaster
g is the acceleration due to gravity (approximately 9.8 m/s^2)
r is the radius of the loop
N is the apparent weight of the student
m is the mass of the student