Answer:
h >5/2r
Explanation:
This problem involves the application of the concepts of force and the work-energy theorem.
The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.
Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)
So from newton's second law of motion,
W – N = mv²/r
N = normal force = 0
W = mg
mg = ma = mv²/r
mg = mv²/r
v²= rg
v = √(rg)
The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop
So
ΔPE = ΔKE = 1/2mv²
The height at the roller coaster starts is usually higher than the top of the loop by design. So
ΔPE =mgh - mg×2r = mg(h – 2r)
2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.
In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.
So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.
ΔPE > ΔKE
mg(h–2r) > 1/2mv²
g(h–2r) > 1/2(√(rg))²
g(h–2r) > 1/2×rg
h–2r > 1/2×r
h > 2r + 1/2r
h > 5/2r
As per the question the initial speed of the car [ u] is 42 m/s.
The car applied its brake and comes to rest after 5.5 second.
The final velocity [v] of the car will be zero.
From the equation of kinematics we know that
[ here a stands for acceleration]
Here a is taken negative as it the car is decelerating uniformly.
We are asked to calculate the stopping distance .
From equation of kinematics we know that
[here S is the distance]
![= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m](https://tex.z-dn.net/?f=%3D%2042%2A5.5%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5B-7.64%5D%20%5B5.5%5D%5E2%20m)
[ans]
Yes it can because it had lots of force
One reason is that when you have been out in the cold, your hands feet and exposed features of your face will take time to recover as the blood circulation improves and supplied warm blood to capillaries. So the relatively warm room you enter will not immediately feel warm until the blood has regained its normal circulation. Other factors are that windows are cooled from the outside and condensation forms on the inside because of moisture in the air. For this condensation to evaporate requires heat, which will be extracted from the room and the air near the windows will be cooled. The cold air will descend and form a draught at floor level and this will tend to make the room cooler.
The color white is what you'd see when every color of light is combined.
