Answer:
[tex]3.6 \times {10}^{5}J[/tex]
Explanation:
From the question, mass(m)=2200kg, unitial velocity(u)=18m/s and final velocity(v)=0m/s
We can calculate the work done to bring the car to a stop from the relation;
[tex]W = F \times S........eqn(1)[/tex],where
W=Work done
F=Force
S=distance
Also,
[tex]F = m \times a............eqn(2)[/tex]
Putting eqn(2) into equn(3) we obatin
[tex]W = m \times a \times S......eqn(3)[/tex]
From the equation of motion;
[tex]a= \frac{v - u}{t}[/tex]
and
[tex]S = (\frac{u + v}{2})t[/tex]
Substituting these into eqn(3), we obtain;
[tex] W =m \times ( \frac{v - u}{t}) \times ( \frac{u + v}{2})t[/tex]
[tex] \implies W=m \times ( v - u) \times (u + v)\times\frac{t}{t} \times \frac{1}{2} [/tex]
[tex]\implies W=m \times ( v - u\times (u + v)\times \frac{1}{2} [/tex]
Substituting the values of m,u and v into the equation, we obtain.
[tex]\implies W=2200 \times ( 0 - 18) (18+ 0)\times \frac{1}{2} [/tex]
Simplifying, we obtain;
[tex]\implies W=1100 \times - 18 \times 18[/tex]
[tex]\implies W= - 356400 = - 3.564 \times {10}^{5} [/tex]
NB: The negative sign indicates that the car decelerated to a stop.
Hence the Work done on the car is
[tex]3.6 \times {10}^{5}J[/tex]
The work done on the car will be "3.6 × 10⁵ J".
According to the question,
Mass, m = 2200 kg
Initial velocity, u = 18 m/s
Final velocity, v = 0 m/s
As we know,
→ Work done (W) = Force(F) × Distance(S) ...(equation 1)
or,
→ Force(F) = Mass(m) × Acceleration(a) ...(equation 2)
From "equation 1" and "equation 2", we get
→ Work done(W) = m × a × S ...(equation 3)
By using Equation of motion,
a = ([tex]\frac{v-u}{t}[/tex])
and,
S = ([tex]\frac{v+u}{2}[/tex])t
Substituting "a" and "S" in "equation 3", we get
→ W = m × ([tex]\frac{v-u}{t}[/tex]) × ([tex]\frac{v+u}{2}[/tex])t
= m × (v-u) × (v+u) × [tex]\frac{t}{t}[/tex] × [tex]\frac{1}{2}[/tex]
= m × (v-u) × (v+u) × [tex]\frac{1}{2}[/tex]
By substituting the values,
= 2200 × (0-18) (0+18) × [tex]\frac{1}{2}[/tex]
= 1100 × (-18)(18)
= -3.564 × 10⁵ or,
= 3.6 × 10⁵ J
Thus the response above is appropriate.
Find out more information about work done here:
https://brainly.com/question/25573309
