Question

A car is driving along at 65 mph up a hill inclined at a 10◦ angle above horizontal when suddenly the engine dies. How high up the hill does it continue to coast? No friction or drag.

same problem as the previous, but the drag force acts like a frictional force with a coefficient of µ = 0.2. How high does the car go this time?

Answer 1

Initial velocity u = 65 mph = 65 miles / hr

= 65 (1609 m) / 3600s

= 29.05 m/s

Required height h = ?

from law of conservation energy , (1/2)mv ² = mgh

h = v ² / 2g

= 29.05 ² /(2x9.8)

= 43 m

(b).Frictional force = mg cos 10

Accleration a = -[g cos 10 + g sin 10 ]

= -g [ cos 10 + sin 10 ]

= -9.8(0.37)

= - 3.631 m/s ²

Final veklocity v = 0

from the relation v ² - u ² = 2aS

29.05 ² - 0 ² = 2(-3.631) S

From this S = 116.17 m

We know sin 10 = H / S

From this required height H = S sin10

= 20.17 m