Answer:
Compression in the spring, x = 0.20 m
Explanation:
Given that,
Spring constant of the spring, k = 490 N/m
Mass of the block, m = 5 kg
To find,
Compression in the spring.
Solution,
Since the block is suddenly dropped on the spring gravitational potential energy of block converts into elastic potential energy of spring. Its expression is given by :
Where
x is the compression in the spring
x = 0.20 m
So, the compression in the spring due to block is 0.20 meters.
Answer: Tangential Velocity
The tangential velocity is defined as the angular velocity
by the radius
of circular motion. As shown below:
Its name is due to the fact that this linear velocity vector is always tangent to the trajectory and is the distance traveled by a body or object in a circular movement in a period of time.
1. -23.2 J
The gravitational potential energy of the ball is given by
where
m = 1.1 kg is the mass of the ball
g = 9.8 m/s^2 is the acceleration of gravity
h is the height of the ball, relative to the reference point chosen
In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is
h = -2.16 m
And the gravitational potential energy is
2. 41.1 J
Again, the gravitational potential energy of the ball is given by
In this part of the problem, the reference point is the floor.
The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:
h = 5.97 m - 2.16 m = 3.81 m
And so the gravitational potential energy of the ball relative to the floor is
3. 0 J
As before, the gravitational potential energy of the ball is given by
Here the reference point is a point at the same elevation of the ball.
This means that the heigth of the ball relative to that point is zero:
h = 0 m
And so the gravitational potential energy is
Hi there!
a.
We know that:
Begin by determining the forces in the vertical direction:
W = weight of traffic light
T₁sinθ = vertical component of T₁
T₂sinθ = vertical component of T₂
b.
The ropes provide a horizontal force:
T₁cosθ = Horizontal component of T1
T₂cosθ = Horizontal component of T2
Thus:
0 = T₁cosθ - T₂cosθ
T₁cosθ = T₂cosθ
T₁ = T₂
c.
Since the angles for both ropes are the same, we can say that:
T₁ = T₂
Sum the forces:
ΣFy = T₁sinθ + T₁sinθ - W = 0
2T₁sinθ = W
d.
Now, we can begin by solving for the tensions:
2T₁sinθ = W