Question

how much work must be done by the brakes to bring the bike rider to the stop? how far does the bicycle travel if it takes 3.5 seconds to come to rest? what is the magnitude of The breaking Force?

Answer 1

Answer:

a) The workdone to bring the rider and bicycle to a stop = -4900J

b) △x = 21 m

c) Magnitude of the braki

Step-by-step explanation:

The total mass of the rider and the bicycle, m = 60 kg + 7.8 kg

m = 67.8 kg

Speed, v = 12 m/s

The initial kinetic energy of the rider and the bicycle is:

KE = 0.5 mv²

KE = 0.5 x 67.8 x 12²

KE = 4881.6 J

KE = 4900 J (to 2 significant figures)

Since both of them are brought to a stop, the final kinetic energy = 0 J

The workdone to bring the rider and bicycle to a stop = Final Kinetic Energy - Initial Kinetic Energy

The workdone to bring the rider and bicycle to a stop = 0 - 4900

The workdone to bring the rider and bicycle to a stop = -4900J

b) The acceleration is calculated as:

`\begin{gathered} a=(v-u)/(t) \\ \\ a=(0-12)/(3.5) \\ \\ a=-3.43\text{ m/s}^2 \end{gathered}`

Calculate the distance using the equation of motion

`\begin{gathered} v^2=u^2+2aS \\ \\ 0^2=12^2+2(-3.43)S \\ \\ 2(3.43)S=144 \\ \\ 6.86S=144 \\ \\ S=(144)/(6.86) \\ \\ S=21\text{ m} \\ \end{gathered}`

The bicycle travels a distance of 21 m

△x = 21 m

c) The magnitude of the braking force:

`\begin{gathered} F=(|W|)/(d) \\ \\ F=(4900)/(21) \\ \\ F=230\text{ N \lparen to 2 significant figures\rparen} \end{gathered}`