How much work must be done on a 10 kg bicycle to increase its speed from 2 ms to 3 ms?

Answer 1

30 J

Answer 2

The bicycle's initial kinetic energy is [ ½ M Vi2 ], and its final kinetic energy is [ ½ M Vf2 ].
The difference in energy is the work that somebody from the outside has to put into it.

The difference is [ (final kinetic energy) minus (initial kinetic energy) ]

[ ½ M Vf2 ] minus [½ M Vi2 ]

or [ ½ M ] times [ Vf2 - Vi2 ]

=[ 5 ] x [ 102 - 52 ]

= [ 5 ] x [ 100 - 25 ]

= [ 5 ] x [ 75 ]

= 375 joules

Answer 3

The bicycle's initial kinetic energy is [ ½ M Vi2 ], and its final kinetic energy is [ ½ M Vf2 ].

The difference in energy is the work that somebody from the outside has to put into it.

The difference is [ (final kinetic energy) minus (initial kinetic energy) ]

[ ½ M Vf2 ] minus [½ M Vi2 ]

or [ ½ M ] times [ Vf2 - Vi2 ] = [ 2.5 ] x [ 152 - 102 ]

= [ 2.5 ] x [ 225 - 100 ] = [ 2.5 ] x [ 125 ]

= 312.5 joules

Answer 4

Kinetic Energy = 1/2 m V2

At 2 m/s, the bicycle's KE is (1/2 x 10 x 4) = 20 joules.

At 3 m/s, its KE is (1/2 x 10 x 9) = 45 joules.

The difference in kinetic energy at the higher speed is (45 - 20) = 25 joules.

That's the energy (work) that has to come from somewhere in order to achieve

the higher speed.

Check:

The ratio of speeds is 3/2 = 1.5 . Since kinetic energy is proportional to the square of the speed,

we need the square of the ratio to check. It's (1.5)2 = 2.25 .

The ratio of the kinetic energies that we calculated at the two speeds is (45/20) = 2.25 .

That's good enough for us. Check ! and mate

Answer 5

100 J

Answer 6

2 Nand the output force 4 N is it’s mechanical advantage

Answer 7

1000