I'm not quite sure what happens to Fay so I didn't finish but hope it helps
An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.
1 answer:
You might be interested in
1) 5.5 N
When the ball is at the bottom of the circle, the equation of the forces is the following:
where
T is the tension in the string, which points upward
mg is the weight of the string, which points downward, with
m = 0.158 kg being the mass of the ball
g = 9.8 m/s^2 being the acceleration due to gravity
is the centripetal force, which points upward, with
v = 5.22 m/s being the speed of the ball
R = 1.1 m being the radius of the circular trajectory
Substituting numbers and re-arranging the formula, we find T:
2) 3.9 N
When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:
And by substituting the numerical values, we find
3) 2.3 N
When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is
And substituting the numerical values and re-arranging it, we find
4) 3.3 m/s
The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:
and re-arranging the equation, we find
Answer:
a) Em₀ = 42.96 104 J
, b) = -2.49 105 J
, c) vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
Em = K + U
a) Let's look for the initial mechanical energy
Em₀ = K + U
Em₀ = ½ m v2 + mg and
Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
Em₀ = 36 104 + 6.96 104
Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
= Em₂ -Em₀
Em₂ = K + U
Em₂ = ½ m v₂² + m g y₂
Em₂ = ½ 50 85 2 + 50 9.8 427
Em₂ = 180.625 + 2.09 105
Em₂ = 1,806 105 J
= Em₂ -Em₀
= 1,806 105 - 4,296 105
= -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
= ΔEm
= Emf - Emo
-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
½ 50.0 vf² = 0.561
vf = √ 0.561 25
vf = 3.75 m / s