Answer:
68.585m/sec , 779.1 N
Explanation:
To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.
From centripetal motion.
F =( mv^2)/2
But since we are dealing with weightlessness
r = 480m
g = 9.8m/s^2
M also cancels, so forget M.
V^2 = Fr
V = √ Fr
V =√ (9.8 x 480) = 4704
= 68.585m/sec.
b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960
= 4.9m/sec^2.
Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)
159kg × ( 9.8-4.9)
159kg × 4.9
= 779.1N
A) The speed of the scooter at which the driver will feel weightlessness is;
v = 68.586 m/s
B) The apparent weight of both the driver and the scooter at the top of the hill is;
F_net = 779.1 N
We are given;
Mass; m = 159 kg
Radius; r = 480 m
A) Since it's motion about a circular hill, it means we are dealing with centripetal force.
Formula for centripetal force is given as;
F = mv²/r
Now, we want to find the speed of the scooter if the driver feels weightlessness.
This means that the centripetal force would be equal to the gravitational force.
Thus;
mg = mv²/r
m will cancel out to give;
v²/r = g
v² = gr
v = √(gr)
v = √(9.8 × 480)
v = √4704
v = 68.586 m/s
B) Now, he is travelling with speed of;
v = 68.586 m/s
And the radius is 2r
Let's first find the centripetal acceleration from the formula; α = v²/r
Thus; α = 4704/(2 × 480)
α = 4.9 m/s²
Now, since he has encountered a hill with a radius of 2r up the slope, it means that the apparent weight will now be;
F_app = m(g - α)
F_net = 159(9.8 - 4.9)
F_net = 779.1 N
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