Respuesta :
Answer:
The velocity of car is 2.19 [tex]\frac{m}{s}[/tex] and horizontal force exerted on the car is 229.5 N
Explanation:
Given :
Mass of car [tex]m = 2.20 \times 10^{3}[/tex] Kg
Work done by car [tex]W = 5280[/tex] J
Displacement [tex]d = 23[/tex] m
From work energy theorem,
[tex]W = \Delta K[/tex]
But here final velocity is zero so we write,
[tex]W = \frac{1}{2} mv^{2}[/tex]
Find velocity,
[tex]v = \sqrt{\frac{2W}{m} }[/tex]
[tex]v = \sqrt{\frac{2 \times 5280}{2200} }[/tex]
[tex]v = 2.19[/tex] [tex]\frac{m}{s}[/tex]
Now, horizontal force exerted on the car,
[tex]W = F \times d \cos \theta[/tex]
[tex]F = \frac{W}{d}[/tex] (∵ [tex]\cos \theta = 1[/tex] )
[tex]F = \frac{5280}{23}[/tex]
[tex]F = 229.5[/tex] N
Therefore, the horizontal force exerted on the car 229.5 N
The speed of the car "v" is 2.19m/s
The horizontal force exerted on the car is 229.57N
The work done on the car is equal to the kinetic energy as shown:
W = KE
Since kinetic energy is expressed as [tex]KE=\frac{1}{2}mv^2\\[/tex]
[tex]W=0.5mv^2\\v^2=\frac{W}{0.5m}\\v^2=\frac{5280}{0.5(2200)}\\v^2=\frac{5280}{1100}\\v^2= 4.8\\v=\sqrt{4.8}\\v= 2.19m/s[/tex]
Hence the speed of the car "v" is 2.19m/s
The horizontal force exerted on the car is expressed as;
[tex]W=Fd\\F =\frac{W}{d}\\F=\frac{5280}{23}\\F= 229.57N[/tex]
Hence the horizontal force exerted on the car is 229.57N
Learn more on workdone here: https://brainly.com/question/9566221
