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:
the can's kinetic energy is 0.42 J
Explanation:
given information:
Mass, m = 460 g = 0.46 kg
diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m
velocity, v = 1.1 m/s
the kinetic energy of the can is the total of kinetic energy of the translation and rotational.
KE = 
I ω^2 + 
where
I = 
and ω = 
thus,
KE = ()^2 +
= 
+
= 
+
= 
= 
= 0.42 J
Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
Answer:
0.5m/s^2
Explanation:
We can use the formula [ F = ma ] but solve for "a" since that is what we are looking for.
F = ma
F/m = a
We know the net force and mass so substitute those values and simplify.
500/1000 = 0.5m/s^2
Best of Luck!
