It is friction that provides the force for a car to acceleate, so for high-perfo

Question

It is friction that provides the force for a car to acceleate, so for high-performance cars the factor that limits acceleration isn't the engine; it's the tires.

1. For typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 50 mph? Suppose static friction coefficient: 1.00, kinetic friction coefficient: 0.80, and rolling friction coefficient: 0.02.

2. A fisherman has caught a very large, 5.0 kg fish from a dock that is 2.0 m above the water. He is using lightweight fishing line that will break under a tension of 60N or more. He is eager to tget the fish to the dock in the shortest possible time.

a. If the fish is at rest at the water's surface, what's the least amount of time in which the fisherman can raise the fish to the dock without losing it?

3. Riders on the Tower of Doom, an amusement park ride, experience 2.0 s of free fall, after which they are slowed to a stop in 0.50 s.

a. What is a 65 kg rider's apparent weight as the ride is coming to rest?

b. By what facotr does this exceed her actual weight?

4. An impala is an African antelope capable of remarkable vertical leap. In one recorded leap, a 45 kg impala went into a deep crouch, pushed straight up for 0.21 s, and reached a height of 2.5 m above the ground.

a. To achieve this vertical leap, with what force did the impala push down on the ground?

b. What is the ratio of this force to the antelope's weight?

If you could show your steps to how you did these problems, it would be much appreciated!

Explanation / Answer

Number 1)

Apply F = ma

The force is from friction

Ff = uFn

umg = ma

ug = a

(1)(9.8) = a

a = 9.8 m/s2

50 mph converts to 23.352 m/s

Apply vf = vo + at

23.352 = 0 + (9.8)(t)

t = 2.38 s

Number 2)

From Newton's Second Law...

60 = mg + ma

60 = (5)(9.8) + (5)(a)

a = 2.2 m/s2

Then apply d = vot + .5at2

2 = 0 + (.5)(2.2)(t2)

t = 1.35 sec

Number 3)

Part A)

We need to first find the velocity after the 2 sec

vf = vo + at

vf = 0 + 9.8(2)

vf = 19.6 m/s

Then apply vf = vo + at for the stop...

0 = 19.6 + a(.5)

a = 39.2 m/s2

F = ma + mg

F = (65)(39.2) + (65)(9.8)

F = 3185 N

Part B)

3185/9.8 = 325 g's

Number 4)

Part A)

We first need the speed of leaving the ground

Apply vf2 = vo2 + 2ad

0 = vo2 + 2(9.8)(2.5)

vo = 7 m/s

Then apply vf = vo + at - during the crouch

7 = 0 + a(.21)

a = 33.3 m/s2

Then F = ma

F = (45)(33.3)

F = 1500 N

Part B)

Take 1500/(45)(9.8)

Ratio = 3.4

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