Answer:
Explanation:
Work done = ∫Fdx
= ∫(cx-3.00x²) dx
[ c x² / 2 - 3 x³ / 3 ]₀²
= change in kinetic energy
= 11-20
= - 9 J
[ c x² / 2 - x³ ]₀² = - 9
c x 2² / 2 - 2³ = -9
2c - 8 = -9
2c = -1
c = - 1/2
Answer:
x = 1.6 + 1.7 t^2 omitting signs
a) at t = 0 x = 1.6 m
b) V = d x / d t = 3.4 t
at t = 0 V = 0
c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)
d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)
Σf = m a
Σf = m v^2 / r
Σf = 52 8^2 / 1.6
Σf = 2080 N
From an energy balance, we can use this formula to solve for the angular speed of the chimney
ω^2 = 3g / h sin θ
Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s
The formula for radial acceleration is:
a = rω^2
So,
a = 53.2 (0.987) = 52.494 /s^2
The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887
The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2
If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67°
Answer:
1.2 kg
Explanation:
Let UP ramp be the positive direction
F = ma
T - Wt || - Ff = m(0)
mg - Μgsinθ - μΜgcosθ = 0
m(9.8) - 13sin35 - 0.36(13)cos35 = 0
m = 13(sin35 + 0.36cos35) / 9.8
m = 1.15205... ≈ 1.2 kg