Question

You’re in the Nevada dessert and your 1800 kg car suddenly runs out of gas. You are going 25 m/s downhill towards a valley in the road, but luckily there is a gas station at the top of the hill on the other side of the valley. You throw your manual transmission car into neutral to coast down the hill and do a little calculation to pass the stressful time. Assume all forms of friction are negligible. (At present your car is at the 10m height of hill and the gas station is 15m height of the hill)
a) How fast will your car be going when it’s at the bottom of the hill? Solve using energy methods. b) Will you be able to get to the gas station? If so, how fast will you be going? If not, how far up the hill did you make it? c) If you model the bottom of the hill as an arc of a circle with radius 5 m, draw the free-body diagram for the car at the bottom of the hill. d) What is the normal force on your car from the road at the bottom of the hill?

Answer 1

Answer:

Explanation:

1 )  Total mechanical energy of the car at the height of 10 m

1/2 mv² + mgh

.5 x 1800 x 25² + 1800  x 9.8 x 10 m

= 562500 + 176400

738900 J  

If v be the velocity at the bottom ,

Total energy of the car at the bottom

1/2 m v² + 0

Applying conservation of energy

1/2 mv² = 738900

.5 x 1800 v² = 738900

v = 28.65 m /s

Energy required by car to ascend height of 15 m

1800 x 9.8 x 15

= 264600 J

b )

This energy is more that total energy of the car at the top that is 738900 J

so car can easily reach gas station .

If V be the velocity at the gas point

Total energy at the gas point

1/2 m V² +  264600

Applying conservation of energy

1/2 m V² +  264600   =738900

.5 x 1800 x V² = 474300

V = 22.95 m / s

d ) If R be the normal reaction at the bottom

net force

R - mg = m v² / r

R = m ( g +  v² / r )

1800 ( 9.8 + 28.65² / 5 )

R = 313136 N