Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.
The answer is either C or D..
<span>The formulas are,
v1d1² = v2d2² ........ (1)
h = (v2²-v1²)/2g ...... (2)
Given that,
v1 = 1.71 m/s
we assume that the stream has decreased by a factor
d2 =0.805d1
then,
v1d1² = v2 (0.805d1)²
cancelled both side d1² then we get,
v1 = v2 (0.805)²
v1 = v2 (0.648025)
Sub v1 = 1.71,
1.71 = v2 (0.648025)
v2 = 1.71/0.648025
v2 = 2.638787083831642
v2 = 2.64 m/s
The vertical distance formula,
h = (v2²-v1²)/2g
We know that value of gravity constant is 9.8 m/s²
h = {(2.64)² - (1.71)²)/2(9.8)
h = {(6.9696) - (2.9241)}/19.6
h = (4.0455)/19.6
h = 0.2064030612244898
h = 0.21 cm
Therefore, the vertical distance h = 0.21 cm.</span>
