Answer:
Explanation:
As the disc is unrolling from the thread then at any moment of the time
We have force equation as
also by torque equation we can say
Now we have
Also from above equation the tension force in the string is
Answer:195 J
Explanation:
Given
mass of ball 
ball leaves the hand with 
maximum height reached by ball 
Initial Mechanical energy when ball just leaves the hand
considering hand to be datum so h_1=0[/tex]
so Potential energy at ground is zero
Mechanical Energy at highest point
at highest Point velocity is zero
Decrease in Mechanical energy
Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
The force of F=10 N produces an extension of
on the string, so the spring constant is equal to
Then the string is stretched by
. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
Answer:
El peso del cartel es 397,97 N
Explanation:
La tensión dada en cada segmento del cable = 2000 N
El desplazamiento vertical del cable = 50 cm = 0,5 m
La distancia entre los polos = 10 m
La posición del letrero en el cable = En el medio = 5
El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °
El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero
El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N
El peso del signo = 397,97 N.
