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?

Question

Alberto Leach

Physics

23 June, 11:20

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?

+5

Answers (1)

Know the Answer?

Maia Robbins · Answer 1

F=m*a=>a=F/m=10000/1267=7.89 m/s²

d=v₀t+a't²/2394.6=112.5*a'=>a'=3.5

a-a'=7.89-3.5=4.39 m/s²

This difference causes friction forces

We apply the second principle of dynamics: vector: F + N + G + Ff = ma (vector vectors, I can not here)

Scalar: Ox: F-Ff = ma

Oy N-mg = 0

Ff = - ma + F = -1267*7.89+10000=-8869+10000=1131 N

This frictional force (Ff) is opposite to the traction (F)