21
A bicyclist was moving at a rate of 5 m/s, and then sped up to 9 m/s. If the
cyclist has a mass of 130 kg, how much work was needed to increase his"
velocity? (Hint: Use the work-kinetic energy theorem.)
O
A. 1625 J
O
B. 6890
O C. 5265 J
O D. 3640 J

The answer for the following problem is mentioned below.

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 () = × m × (u²)

initial kinetic energy () =   × 130 × ( 5 ×5 )

initial kinetic energy () = 65  × 25 =1625 J

final kinetic energy() =   × m × (v²)

final kinetic energy() =   × 130 × ( 9×9 )

final kinetic energy() = 81 × 65

final kinetic energy() = 5265 J

Work done (W) =ΔK.E =() - ()

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

Work done (W) = 3640 J

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

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )

explanation:

wheres the pic