Answer:
[tex]\mathtt{\Delta K.E = 10500 \ J}[/tex]
Step-by-step explanation:
Given that:
the expression for the change in kinetic energy = [tex]\dfrac{1}{2} \times 350 \times ( 122 -62)[/tex]
Recall that
Kinetic energy K.E = [tex]\dfrac{1}{2}mv^2[/tex]
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
[tex]\Delta K.E = K.E_2 - K.E_1[/tex]
[tex]\Delta K.E =\dfrac{1}{2}mv^2_2- \dfrac{1}{2}mv^2_1[/tex]
[tex]\Delta K.E =\dfrac{1}{2}m(v^2_2-v^2_1)[/tex]
where;
m = 350
[tex]v_2[/tex] = 122
[tex]v_1[/tex]= 62
[tex]\Delta K.E =\dfrac{1}{2} \times 350 \times (122-62)[/tex]
[tex]\Delta K.E =\dfrac{1}{2} \times 350 \times (60)[/tex]
[tex]\Delta K.E = 350 \times 30[/tex]
[tex]\mathtt{\Delta K.E = 10500 \ J}[/tex]
