Question

21

Answer 1

The answer for the following problem is mentioned below.

• Therefore the work done to increase the velocity is 3640 J.

Explanation:

Given:

initial speed (u) =5 m/s

final speed (v) = 9 m/s

mass of the cyclist (m) =130 kg

To solve:

Wok needed to increase the velocity (W)

We know;

initial kinetic energy ($KE_{1}$) = $\frac{1}{2}$ × m × (u²)

initial kinetic energy ($KE_{1}$) =  $\frac{1}{2}$ × 130 × ( 5 ×5 )

initial kinetic energy ($KE_{1}$) = 65  × 25 =1625 J

final kinetic energy($KE_{2}$) =   $\frac{1}{2}$ × m × (v²)

final kinetic energy($KE_{2}$) =  $\frac{1}{2}$ × 130 × ( 9×9 )

final kinetic energy($KE_{2}$) = 81 × 65

final kinetic energy($KE_{2}$) = 5265 J

Work done (W) =ΔK.E =($KE_{2}$) - ($KE_{1}$)

Work done (W) = 5265 - 1625 =3640 J

Work done (W) = 3640 J

Therefore the work done to increase the velocity is 3640 J.