Answer:
The work done is 354.1032 Joules
Explanation:
To answer the question we note that the amount of work done is equivalent to the reduction in the kinetic energy of the bicyclist and bicycle.
Therefore, initial kinetic energy. KE[tex]_i[/tex] of the system is given by the following relation;
[tex]KE_i=\frac{1}{2} mv_{1}^{2}[/tex]
Final kinetic energy [tex]KE_f = \frac{1}{2} mv_{2}^{2}[/tex]
Where:
v₁ = Initial velocity = 3.95 m/s
v₂ = Final velocity = 1.31 m/s
m = Mass of the bicyclist and bicycle = 51.0 kg
Change in kinetic energy ΔKE;
[tex]\Delta KE = KE_f - KE_i[/tex]
[tex]\Delta KE =\frac{1}{2} mv_{2}^{2}-\frac{1}{2} mv_{1}^{2} = \frac{1}{2} m(v_{2}^{2}-v_{1}^{2}) = \frac{1}{2} \times 51 \times (1.31^{2}-3.95^{2}) = -354.1032 \ J[/tex]
Therefore, work done = -ΔKE = 354.1032 Joules.
