Answer:
0.466 (3 sig. fig.)
Explanation:
Frictional force acting on the box = 5.00×10^2xsin25
Normal force acting on the box = 5.00×10^2xcos25
coefficient of friction = 0.466 (3 sig. fig.)
To find a general equilibrium point for a spring based on the hook law, it is possible to start from the following premise:
Hook's law is given by:

Where,
k= Spring Constant
Change in Length
F = Force
When there is a Mass we have two force acting at the System:
W= mg
Where W is the force product of the weigth. Then the force net can be defined as,

But we have a system in equilibrium, so

We find the equilibrium for any location when

That Energy Cannot Be Created Nor Destroyed
Answer:
v = (10 i ^ + 0j ^) m / s, a = (0i ^ - 9.8 j ^) m / s²
Explanation:
This is a missile throwing exercise.
On the x axis there is no acceleration so the velocity on the x axis is constant
v₀ₓ = 10 m / s
On the y-axis velocity is affected by the acceleration of gravity, let's use the equation
v_y =
- g t
at the highest point of the trajectory the vertical speed must be zero
v_y = 0
therefore the velocity of the body is
v = (10 i ^ + 0j ^) m / s
the acceleration is
a = (0 i ^ - g j⁾
a = (0i ^ - 9.8 j ^) m / s²