Respuesta :
Using the theorem of Work and Energy:
m = mass of the car = 1167 Kg
V = Final Velocity of the car
v = initial velocity of the car = 10 m/s
Net Work Done = 1/2(mV^2 - mv^2)
4 x 10^5 = 0.5 x 1167 (V^2 - 10^2)
V= (458350/583.5)^1/2
= 28 m/s
Answer:
final speed = 28 m/s
Explanation:
As per work energy theorem we know that
Work done on the object = change in kinetic energy
so here we have
[tex]W = KE_f - KE_i[/tex]
now we will have
[tex]W = 4.00 \times 10^5 J[/tex]
also we know that
mass = 1167 kg
and initial speed is given as
[tex]v_i = 10.0 m/s[/tex]
now from above work energy equation we will have
[tex]4.00 \times 10^5 = \frac{1}{2}1167(v_f^2) - \frac{1}{2}1167(10^2)[/tex]
[tex]4.00 \times 10^5 = \frac{1}{2}1167 v_f^2 - 58350[/tex]
[tex]\frac{1}{2}(1167)v_f^2 = 4.00 \times 10^5 + 58350[/tex]
[tex]v_f = 28 m/s[/tex]
