Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.
Look at the formula for the gravitational force:
F = G m₁ m₂ / R² .
If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to m₂ ... mass of the box,
and you can write a simple proportion:
(6.1 N) / (2.5 kg) = (F) / (1 kg)
Cross-multiply: (6.1 N) (1 kg) = (F) (2.5 kg)
Divide each side by (2.5 kg): F = (6.1N) x (1 kg) / (2.5 kg) = 2.44 N .
Answer:
u=36.8m/s
Explanation:
because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations
u^2=v^2-2ā*s. where:
u^2 stands for intial velocity
v^2 stands for final velocity
since the cougar skidded to a complete stop the final velocity is zero.
u^2=v^2-2ā*s
u^2=(0)^2 -2(-2.87 m/s^2)*236 m
u^2=0+5.74m/s^2* 236m
u^2=1354.64m^2/s^2
u=√1354.64m^2/s^2
u=36.8m/s (approximate value)
when ever the acceleration is constant you can use one of the following equation to find the required value.
1. v = u + at. (no s)
2. s= 1/2(u+v)t. (no ā)
3. s=ut + 1/2at^2. ( no v)
4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)
Answer:
ΔU = 2 mg h
Explanation:
In a spring mass system the potential energy is U = m g h
where h is measured from the equilibrium point of the spring
the potential energy at the highest point is
U₁ = m g h
the potential energy at the lowest point is
U₂ = m g (-h)
instead in this energy it is
ΔU = 2 mg h
In this two points the kinetic energy is zero, but there is elastic potential energy that has the same value in the two points, so its change is zero
