Answer:
x₂=2×1
Explanation:
According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;
mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.
mgx=(kx)²/2
x=2mg/k----------------compression when the object is at rest
However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy
Thus, if kx²=mv² then
v=x *√(k/m) ----------------where v=0
<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁
Answer
Data provided:
mass of the block = 200 g = 0.2 Kg
Velocity at A = 0 m/s
Velocity at B = 8 m/s
distance of slide = 10 m
height of the block = 4 m
calculation for the block's potential energy
P = m g h
P = 0.2 x 9.8 x 4
P = 7.84 J
kinetic energy calculated as



Work done = P - KE
work = 7.84 - 6.14
work = 1.7 J
b) using the formula v² = u² + 2 a s
0 = 8² - 2 x a x 10
a = 3.2 m/s²
ma - μ mg = 0



Answer:
A) and B) are valid.
Explanation:
When an object remains at rest, it is indicative that no net force acts upon it.
The downward gravitational force from Earth must be counterbalanced by an upward force of equal magnitude in order to maintain rest.
This upward force is provided by the normal force, which adjusts to satisfy Newton’s 2nd Law and is always perpendicular to the surface supporting the object (in this instance, the ground).
At the molecular level, this normal force comes from the ground's bonded molecules acting like tiny springs, compressed by the object’s molecules, providing an upward restorative force.
Thus, statements A) and B) are true.
Response:
(a) 104 N
(b) 52 N
Clarification:
Provided Information
Incline angle of the ramp: 20°
F forms a 30° angle with the ramp
The parallel component of F along the ramp is Fx = 90 N.
The perpendicular component of F is Fy.
(a)
Consider the +x direction pointing up the slope, and the +y direction perpendicular to the ramp's surface.
Using the Pythagorean theorem, decompose F into its x-component:
Fx=Fcos30°
To find F:
F= Fx/cos30°
Insert the value for Fx based on the given info:
Fx=90 N/cos30°
=104 N
(b) Calculate the y-component of r using the Pythagorean theorem:
Fy = Fsin 30°
Substituting for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
<span>Conclusion: The door's weight results in a CCW torque which can be calculated as
Tccw = 145 N*3.13 m / 2.
You require a CW torque that balances this
Tcw = F*2.5 m*sin20</span>