# Energy Question Problem

1. Nov 23, 2013

### neededthings

A cyclist is competing in a race and decides to pass some fellow cyclists who are getting tired. The initial speed of the cyclist is 21 km/h and he has 3.0 * 10^2 W of power. If the bicycle is able to convert 92% of input energy into kinetic energy, how fast will cyclist be travelling after 4.0 s. The combined mass of the cyclist and the bicycle is 78 kg.

The answer is 7.9 m/s. But I dont know how.

2. Nov 23, 2013

### Kishlay

3. Nov 23, 2013

### neededthings

energy= Pt= 3x10^2 * 4 = 1.2 x10^3 J
Useful energy = 1.2 x10^3 *0.92 = 1.104 x10^3
KE= 1/2m( v)^2
root(1.104 x10^3 *2 /78) = change in v = +/-5.32 m/s (3 S.F.). He is accelerating hence +5.32m/s
21km/h = 21000m/60x60s= 5.83 m/s (3 S.F)
Total speed = 5.32+5.83 = 11.15m/s

4. Nov 23, 2013

### Kishlay

useful energy =1.104x10^3 ... right
then change in kinetic energy will be equal to useful energy given in 4 seconds.......
the formula 1/2mv^2, 'V' is the instataneous velocity, not the change in velocity
so we get
change in kinetic energy = useful energy given..
mx(Vfinal2-Vinitial2)/2

5. Nov 23, 2013

### neededthings

6. Nov 23, 2013

### Dick

Whoa, back it up a bit. You've got the energy input correct. What's your initial kinetic energy? Add the energy input to that to get the final kinetic energy. Use that to get a final velocity.

7. Nov 23, 2013

### Kishlay

initial kinetic energy + useful energy given= final kinetic energy

8. Nov 23, 2013

### Kishlay

hope this will make you clear

9. Nov 23, 2013

### neededthings

10. Nov 23, 2013

### neededthings

never mind i got it.

Thanks alot

11. Nov 23, 2013

### Kishlay

he he he ...........
if you have really got it then it is fine, if not we are always to help you.....!!

12. Nov 23, 2013