Answer:
The second one
Explanation:
The second one is joining so it is the one which belongs to X, first one belongs to Z
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
Answer:
Explanation:
The given time is 1 / 4 of the time period
So Time period of oscillation.
= 4 x .4 =1.6 s
When the block reaches back its original position when it came in contact with the spring for the first time , the block and the spring will have maximum
velocity. After that spring starts unstretching , reducing its speed , so block loses contact as its velocity is not reduced .
So required velocity is the maximum velocity of the block while remaining in contact with the spring.
v ( max ) = w A = 1.32 m /s.
Answer:
a) θ = 2500 radians
b) α = 200 rad/s²
Explanation:
Using equations of motion,
θ = (w - w₀)t/2
θ = angle turned through = ?
w = final angular velocity = 1420 rad/s
w₀ = initial angular velocity = 420
t = time taken = 5s
θ = (1420 - 420) × 5/2 = 2500 rads
Again,
w = w₀ + αt
α = angular accelaration = ?
1420 = 420 + 5α
α = 1000/5 = 200 rad/s²
maximum static friction acting on the object will be
plug in all values
So here it means that if applied force is less than or equal to 58.8 N then the object will remain stationary as friction can balance the external force upto this limit of external force
So here it is given that applied force is 20 N
so here object will not move due to this force and it will remain at rest always
due to this applied force
