Answer: The work done is W = 3528J.
Explanation: Work (W) is an amount of energy used to do determined movement. To calculate it, we use the formula, as the movement is horizontal:
W = F.d, where
F is force on the object
d is distance it went through.
The force acting on the crate is due to friction.
= μ .
μ is the kinetic friction
is the normal force
There is no up and down movement, so , with g = 9.8m/s²
= μ . m . g
= 0.9 . 80 . 9.8
= 705.6 N
Substituing into the work formula, we have:
W = . d
W = 705.6 . 5
W = 3528 J
The work done to move the crate across a distance of 5m is 3528 J.
Potential energy due to gravity= mass x gravity x height
What’s the question? I may be able to help
Answer:
223.2
Explanation:
126.3851 + 7.2 + 73.607 + 15.98 = 223.1721
When adding/subtracting, round to the fewest decimal places. 7.2 has 1 number after the decimal point, so we'll round our answer to 223.2.
Answer:
R = 0.237 m
Explanation:
To realize this problem we must calculate the moment of inertia of the wheel formed by a thin circular ring plus the two bars with an axis that passes through its center.
The moments of inertia of the bodies are additive quantities whereby we can add the mounts inertia of the ring and the two bars.
Moment of inertia ring I1 = MR²
Moment of inertia bar I2 = 1/12 ML²
Moment of inertia disk I3 = ½ mR²
Let's calculate the moment of inertia of the wheel
I = I1 + 2 I2
I = MR² + 2 1/12 ML²
The length of the bar is ring diameter
L = 2R
I = 5.65 0.156² + 1/6 9.95 (2 0.156)²
I = 0.1375 + 0.1614
I = 0.2989 kg m²
This is the same moment of inertia of the solid disk,
Disk
I3 = I
I3 = ½ MR²
They give us disk density
ρ = M / V
M = ρ V
M = ρ (pi R² e)
Done is the thickness of the disc, in general it is e= 1 cm = 0.01 m
Let's replace
I3 = ½ ( ρ π R²) R²
I3 = ½ ρ π e R⁴
R⁴ = 2 I3 / ( ρ π e)
R = ( 2 I3 / ( ρ π e)
R⁴ = 2 0.2989 / (5990 π 0.01)
R = 0.237 m