Suppose that 2 J of work are needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. How far (in cm) b
1 answer:
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched
<u>Explanation:</u>
Work, W = 2 J
Initial distance, = 30 cm
Final distance, = 42 cm
Force, F = 30 N
Stretched length, x = ?
We know,
W = 1/2 kΔx²
Δx = 42-30 cm = 12 cm = 0.12 m
2 J = 1/2 k X (0.12)²
k = 277.77 N/m
According to Hooke's law,
F = kx
30 N = 277.77 X x
x = 0.108 m
x = 10.8 cm
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.
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Answer: (a) The bicycle is ahead of the car for 4 s.
(b) The bicycle leads the car by the maximum distance of 55 m.
Explanation:
(a)
Use the equation of the motion to calculate the time taken by the car.
As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .
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As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.
Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.
Calculate the time interval for which the bicycle is ahead of the car.
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(b)
Use the equation motion to calculate the distance covered by the car.
As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .
Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.
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Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.
Calculate the maximum distance covered by the bicycle to lead the car.
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