Question

The change in kinetic energy (in Joules) as a horse accelerates from canter to gallop is found using the expression 12×350×(122−62). Which is the change in kinetic energy?

Answer 1

The answer to your question is: 252000

Explanation:

To get the change in kinetic energy, only do the operations given considering priority.

1.- Parenthesis

2.- Multiplication

12×350×(122−62)

12 x 350 x (60)

252000

Answer 2

The change in kinetic energy is
`10500J`

Explanation:

The expression given is wrong. It should be :

`((1)/(2)).(350).(122-62)` (I)

The equation to calculate the kinetic energy
`(KE)` is :

`KE=((1)/(2))mv^(2)`

Where ''m'' is the mass of the object (in this case it will be the mass of the horse).

Where ''v'' is the speed of the object (in this case it will be the speed of the horse).

The kinetic energy is a scalar quantity.

When the unit of ''m'' is kg and the unit of ''v'' is
`(m)/(s)` (meters per second), the unit of kinetic energy is Joule (J).

To calculate the change in kinetic energy between two instants of time, for example between 1 and 2, we can obtain the following equation :

Δ
`KE_(1,2)` =
`KE_(2)-KE_(1)`

Δ
`KE_(1,2)=((1)/(2))mv_(2)^(2)-((1)/(2))mv_(1)^(2)`

Δ
`KE_(1,2)=((1)/(2))m(v_(2)^(2)-v_(1)^(2))` (II)

If we compare the expression (I) and (II) we find that the horse mass is
`350kg`,
`v_(2)^(2)=122(m^(2))/(s^(2))` and
`v_(1)^(2)=62(m^(2))/(s^(2))`.

Finally, if we solve (I) :

`((1)/(2)).(350)(122-62)=10500J`