Someone help please!!
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Answer:
Check attachment for necessary information
Step-by-step explanation:
At point B. Check attachment for free body diagram.
Where an is the centripetal acceleration and it is given as
an = Vb²/r
Fn = m•Vb²/r
Fn = 6 × Vb²/1.2
Fn = 5Vb²
Applying Newton's second law along the y direction
ΣF = m•ay
ay = 0, since the body is not moving in y direction
N —W = 0
N — WCosβ = 0
N = Fn = 5Vb²
5Vb²—58.86Cosβ = 0
Divide through by 5
Vb² — 11.772Cosβ = 0
Vb² = 11.772Cosβ, equation 1
Applying conservation of energy.
∆K.E(A) + ∆P.E(A) = ∆K.E(B) +∆P.E(B)
½m•Va²— 0+ mg•Ha — 0 = ½m•Vb² — 0 + mg•Hb — 0
½m•Va²+mg•Ha = ½m•Vb² + mg•Hb
From attachment
Ha = 1.13m
Hb = 1.2Cosβ.
Va = 2m/s²
½m•Va²+mg•Ha = ½m•Vb² + mg•Hb
½×6×2² + 6×9.81×1.13 = ½×6×Vb²+6×9.81×1.2Cosβ
12 + 66.512 = 3Vb² + 70.632Cosβ
From equation,.Vb² = 11.772Cosβ
78.512 = 3×11.772Cosβ+70.632Cosβ
78.512 = 35.316Cosβ+70.632Cosβ
78.512 = 105.948Cosβ
Cosβ = 78.512/105.948
Cosβ = 0.7410
β = ArcCos(0.7410)
β = 42.18 °
β = 42.2°
From the attachment, it is notice that,
θ + 20° = β
θ = β — 20°
θ = 42.2 — 20°
θ = 22.2°
The angle θ at which the box
left the smooth circular ramp is 22.2°
b. Using equation free fall motion
∆y = Vby•t + ½gt²
Let get Vb first, from equation 1
Vb² = 11.772Cosβ
Vb² = 11.772Cos42.2
Vb² = 8.721
Vb = √8.722
Vb = 2.95m/s
Now, to get Vby
Vby = VbSinβ
Vby = 2.95Sin42.2
Vby = 1.984 m/s
Then, applying free fall equation at point B
∆y = Vby•t + ½gt²
Hb - 0 = 1.984t + ½ × 9.81t²
1.2Cosβ = 1.984t + 4.905t²
1.2Cos42.2 = 1.984t + 4.905t²
4.905t² + 1.984t —0.889 = 0
Using formula method
t = [-b±√(b²-4ac)]/2a
a = 4.905 b = 1.984 and c = -0.889
t =[-1.984±√(1.984²-4×4.905×-0.889)] / 2×4.905
t = (-1.984±√21.378)/9.81
t = (-1.984± 4.624)/9.81
So,
t = (—1.984 — 4.624)/9.81
t = -0.674s
Or
t = (-1.984 + 4.624)/9.81
t = 2.64/9.81
t = 0.269s
Since time cannot negative, then,
t = 0.269s
Now, we can find the distance "s" by applying range formula, the part of motion is parabola this allow us to use projectile motion
R = Ux • t
s= Vbx × t
s= VbCosβ × t
s= 2.95Cos42.2 × 0.269
s= 0.588m
So, the distance "s" where the box fall into the cart is 0.588m
