Question

The change in kinetic energy (in joules) as a horse accelerates from canter to gallop is found using the expression 1/2 times 350 times (12 2 Minus 6 2

Answer 1

Answer:

$\mathtt{\Delta K.E = 10500 \ J}$

Step-by-step explanation:

Given that:

the expression for the change in kinetic energy = $\dfrac{1}{2} \times 350 \times ( 122 -62)$

Recall that

Kinetic energy K.E = $\dfrac{1}{2}mv^2$

where,

m = mass of the horse

v = velocity of the horse

The change in kinetic energy between  two instant times can be expressed by the relation

$\Delta K.E = K.E_2 - K.E_1$

$\Delta K.E =\dfrac{1}{2}mv^2_2- \dfrac{1}{2}mv^2_1$

$\Delta K.E =\dfrac{1}{2}m(v^2_2-v^2_1)$

where;

m = 350

$v_2$ = 122

$v_1$= 62

$\Delta K.E =\dfrac{1}{2} \times 350 \times (122-62)$

$\Delta K.E =\dfrac{1}{2} \times 350 \times (60)$

$\Delta K.E = 350 \times 30$

$\mathtt{\Delta K.E = 10500 \ J}$