(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that
0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)
Then the car comes to rest over a distance of
<em>d</em> = - <em>v</em>² / (2<em>a</em>)
Doubling the starting speed gives
- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>
so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.
Alternatively, you can explicitly solve for the acceleration, then for the distance:
A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that
0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²
So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that
0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m
(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that
60 J = <em>m g h</em>
where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives
<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg
(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.
