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A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds

[a]What is the expected acceleration of the car from 10,000 N force?

[b]What is the actual acceleration from the observed data of x and t?

[c]What is the difference in accelerations?

[d]What force caused this difference?

[e]What is the magnitude and direction of the force that caused the difference in acceleration?

Respuesta :

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/[tex]s^{2}[/tex]

B. Formula: a=[tex]\frac{V-V_{0} }{t}[/tex] and s=d/t

speed= 394.6/15

s=26.3m/s

a=[tex]\frac{26.3-0}{15}[/tex]

a=1.75m/[tex]s^{2}[/tex]

C. 7.9-1.75=difference of 6.15m/[tex]s^{2}[/tex]

D. The force that most likely caused this difference is friction forces

Answer:

Part a)

[tex]a = 7.89 m/s^2[/tex]

Part b)

[tex]a = 3.51 m/s^2[/tex]

Part c)

[tex]\Deltra a = 4.38 m/s^2[/tex]

Part d)

This difference in acceleration  is due to some frictional force on the surface.

Part e)

[tex]F_f = 5552.8 N[/tex]

Explanation:

Part a)

As we know by newton's II law

[tex]F = ma[/tex]

here we know that

[tex]m = 1267 kg[/tex]

[tex]F = 10,000 N[/tex]

Now we have

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{10,000}{1267}[/tex]

[tex]a = 7.89 m/s^2[/tex]

Part b)

distance covered by the car

[tex]d = 394.6 m[/tex]

t = 15 s

now by kinematics we have

[tex]d = \frac{1}{2}at^2[/tex]

[tex]394.6 = \frac{1}{2}a(15^2)[/tex]

[tex]a = 3.51 m/s^2[/tex]

Part c)

Difference of acceleration is given as

[tex]\Delta a = a_{expected} - a_{real}[/tex]

[tex]\Delta a = 7.89 - 3.51 [/tex]

[tex]\Deltra a = 4.38 m/s^2[/tex]

Part d)

This difference in acceleration  is due to some frictional force on the surface.

Part e)

Now for magnitude of force is given as

[tex]F - F_f = ma[/tex]

[tex]10,000 - F_f = ma[/tex]

[tex]10,000 - F_f = 1267\times 3.51[/tex]

[tex]F_f = 5552.8 N[/tex]

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