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
according to newton's second law , net force on an object is equal to the product of its mass and its acceleration. the formula is given as
F = ma
where F = force , m = mass , a = acceleration
rearranging the equation for acceleration "a" , we get
a = F/m
when force F is kept constant , the acceleration is inversely related to the mass "m". in other words, greater the mass , smaller will be the acceleration and smaller the mass, greater will be the acceleration.
the masses of baseball, basketball, tennis ball and bowling ball can in arranged in increasing order as
tennis ball < baseball < basketball < bowling ball.
since acceleration and mass have inverse relation from the formula , the order of the acceleration will also be reverse.
tennis ball > baseball > basketball > bowling ball.
hence the tennis ball will have the greatest acceleration as it has the smallest mass.
Low blood pressure. The person could faint and have an irregular heartbeat.
Answer:
The work required is -515,872.5 J
Explanation:
Work is defined in physics as the force that is applied to a body to move it from one point to another.
The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.
Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:
where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).
The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:
In this case:
- W=?
- m= 2,145 kg
- v2= 12
- v1= 25
Replacing:
W= -515,872.5 J
<u><em>The work required is -515,872.5 J</em></u>
