Answer:
a) K = 49.5 J b) k = 1378 N / m c) ΔE = 34 J d) μ = 0.399
Explanation:
For this exercise we will use the concepts of energy
a) The initial kinetic energy is
K = ½ m v²
K = ½ 3.96 5²
K = 49.5 J
b) let's use energy conservation
Em₀ = K = ½ m v²
= Ke = ½ k x²
Em₀ =
½ m v² = ½ k x²
k = m v² / x²
k = 3.96 5² / 0.268²
k = 1378 N / m
c) Let's calculate the final energy of the spring
= Ke = ½ k x²
= ½ 1378 0.15²
= 15.5 J
The initial energy is the kinetics of the block
Em₀ = 49.5 J
The lost energy is the difference with the initial
ΔE = - Em₀
ΔE = 15.5 - 49.5
ΔE = - 34 J
the negative sign means that the energy dissipates
d) For this part we use the concept of work
W = F d cos θ = ΔK
In this case the force is the friction force that always opposes displacement, so the angle 180 ° and cos 180 = -1
W = -fr d = ΔK
The force of friction is
fr = μ N
With Newton's second law
N-w = 0
N = W = mg
Let's calculate
-μ mg d = Kf -K₀o
μ = K₀ / mgd
μ = 49.5 / (3.96 9.8 3.20)
μ = 0.399
Divide by 9.8 I think
so <span>1938.77551</span>
Mechanical Advantage is the value which tells us the factor by which the machine is able to convert the force applied to it to its output.
Ideal Mechanical Advantage of wheel and axle is calculated by
IMA = Wheel Radius / Axle Radius
By virtue, the higher the computed IMA the lower the force needed to be exerted, this is the purpose for using machine, lower input force but larger output. (The smalled the axle radius compared to the wheel radius is the better setup.)
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
where;
Fc is the centripetal force
is downward force due to weight of the driver
is upward or normal force on the drive
Therefore, the normal force the seat exerted on the driver is 125 N.
