How does a bicycle rider's energy of motion changes as the gain speed while riding down a hill?
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Answer:
13.78 m/s
Explanation:
Given that:
The mass of the girl & her bicycle = 42 kg
The speed at the top of the hilll= 4 m/s
The height of the hill = 14.3 m
The length of the hill (distance) = 112 m
The frictional force = 20 N
To find the speed at the bottom of the hill; we need to carry out the following processes.
The workdone by gravity = mass × acceleration due to gravity × Δh
= 42 × 9.8 × ( 14.3 - 0 )
= 5885.88 joules
The workdone by the friction = Force × distance = - 20 × 112 (since she is riding down the hill)
The workdone by the friction = -2240 Joules
The initial Kinetic friction = 1/2 mv²
= 1/2 × 42 × 4²
= 336 Joules
The final kinetic energy = Initial Kinetic energy + total work
The final kinetic energy = (336 + 5885.88 - 2240) Joules
The final kinetic energy = 3981.88 Joules
Using the final kinetic energy = 1/2 mv²
3981.88 = 1/2 × 42 × v²
3981.88 = 21 v²
v² = 3981.88/21
v² =189.61
v =
v = 13.78 m/s
Therefore, the speed at the bottom = 13.78 m/s