Given,
The braking force, F=40 N
The initial velocity of the cyclist, u=20 m/s
The final velocity of the cyclist, v=15 m/s
The time duration, t=2 s.
From the equation of motion,
[tex]d=\frac{1}{2}(v+u)t[/tex]Where d is the distance traveled by the cyclist after applying the brake and before acquiring the final velocity.
On substituting the known values,
[tex]\begin{gathered} d=\frac{1}{2}\times(15+20)\times2 \\ =35\text{ m} \end{gathered}[/tex]The work done on the cyclist is given by,
[tex]W=F\times d[/tex]On substituting the known values,
[tex]\begin{gathered} W=40\times35 \\ =1400\text{ J} \end{gathered}[/tex]Therefore the work done on the cyclist is 1400 J.
Thus, the correct answer is option A.
