This problem is looking for the minimum value of μs that is
necessary to achieve the record time. To solve this problem:
Assuming the front wheels are off the ground for the entire
¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².
For a constant acceleration, distance = 402.3
m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2
µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18
Answer:
a)-1.014x
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b)3.296 x
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Explanation:
For Sphere A:
mass 'Ma'= 47kg
xa= 0
For sphere B:
mass 'Mb'= 110kg
xb=3.4m
a)the gravitational potential energy is given by
= -GMaMb/ d
= - 6.67 x
x 47 x 110/ 3.4 => -1.014x
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b) at d= 0.8m (3.4-2.6) and
=-1.014x
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The sum of potential and kinetic energies must be conserved as the energy is conserved.
+
=
+ 
As sphere starts from rest and sphere A is fixed at its place, therefore
is zero
=
+ 
The final potential energy is
= - GMaMb/d
Solving for '
'
=
+ GMaMb/d => -1.014x
+ 6.67 x
x 47 x 110/ 0.8
= 3.296 x
J
Answer:
H = 54.37
Explanation:
given,
lead ball attached at = 1.70 m
rate of revolution = 3 revolution/sec
height above the ground = 2 m

circumference of the circle = 2 π r
= 2 x π x 1.7
= 10.68 m

v = 32.04 m/s
using conservation of energy




H = 54.37
the maximum height reached by the ball is equal to H = 54.37