The second object, the one that had twice the force applied to it, would move twice as far, I believe.
Because the weight of one ball is mg = 147 N, the gravitational force between the two balls is or parts per billion of the weight.
The initial speed of the automobile is 49.84km/hr
<u>Explanation:</u>
Given:
Acceleration, a = 1.77 m/s²
Time, t = 6s
Final speed, v = 88 km/h
v = 88 X 0.278 m/s
v = 24.464 m/s
Initial speed, u = ?
We know,
v = u + at
On substituting the value in the formula we get:
24.464 = u + (1.77 X 6)
24.464 = u + 10.62
u = 24.464 - 10.62 m/s
u = 13.844 m/s
Converting u = 13.844 m/s to km/hr
1 m/s = 3.6 km/hr
13.844 m/s = 13.844 X 3.6 km/hr
u = 49.84 km/hr
Therefore, the initial speed of the automobile is 49.84km/hr
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>