(8) A car starting with a speed v skids to a stop over a distance d, which means the brakes apply an acceleration a such that
0² - v² = 2 a d → a = - v² / (2d)
Then the car comes to rest over a distance of
d = - v² / (2a)
Doubling the starting speed gives
- (2v)² / (2a) = - 4v² / (2a) = 4d
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 a such that
0² - (13.9 m/s)² = 2 a (15 m) → a ≈ -6.43 m/s²
So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance d such that
0² - (27.8 m/s)² = 2 (-6.43 m/s²) d → d ≈ 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 m such that
60 J = m g h
where g = 10 m/s² and h is the height it is lifted, 1.2 m. Solving for m gives
m = (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.