Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm = - Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m
Explanation:
A person standing still for a long time feels tired because the force of gravity acts on our body and puts stress on our muscles. so our muscles need energy to do work and keep body balanced and help to stand upright.
Answer:
E) True. Ball B will go four times as high as ball A because it had four times the initial kinetic energ
Explanation:
To answer the final statements, let's pose the solution of the exercise
Energy is conserved
Initial
Em₀ = K
Em₀ = ½ m v²
Final
Emf = U = mg h
Em₀ = emf
½ m v² = mgh
h = v² / 2g
For ball A
h_A = v² / 2g
For ball B
h_B = (2v)² / 2g
h_B = 4 (v² / 2g) = 4 h_A
Let's review the claims
A) False. The neck acceleration is zero, it has the value of the acceleration of gravity
B) False. Ball B goes higher
C) False has 4 times the gravitational potential energy than ball A
D) False. It goes 4 times higher
E) True.
Fnet=(115+106)-186= 34 N
mass=Force/g= 186N/9.8m/s^2 = 18.98 kg
a=fnet/mass => 34N/18.98kg = 1.79 m/s^2
so A= 1.8m/s^2
Answer:
(a) 29 cm
(b) 43.5 cm
Explanation:
(a) when loop A is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 70 cm mark
-2 N at x
Taking the sum of the torques about B:
∑τ = Iα
(-0.9 N) (50 cm − 70 cm) + (-2 N) (x − 70 cm) = 0
18 Ncm − 2 N (x − 70 cm) = 0
2 N (x − 70 cm) = 18 Ncm
x − 70 cm = 9 cm
x = 79 cm
The distance from the center is |50 cm − 79 cm| = 29 cm.
(b) when loop B is slack, there are three forces acting on the metre rule.
-0.9 N at 50 cm mark
T at 20 cm mark
-2 N at x
Taking the sum of the torques about A:
∑τ = Iα
(-0.9 N) (50 cm − 20 cm) + (-2 N) (x − 20 cm) = 0
-27 Ncm − 2 N (x − 20 cm) = 0
2 N (x − 20 cm) = -27 Ncm
x − 20 cm = -13.5 cm
x = 6.5 cm
The distance from the center is |50 cm − 6.5 cm| = 43.5 cm
