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valentinak56 [21]
3 years ago
10

A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i

s travelling at 14.5m/s and only the friction from the tires on the roadway stop the vehicle. Find the coefficient of friction
Physics
1 answer:
Nastasia [14]3 years ago
6 0

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

v^2=u^2+2as

0^2=14.5^2+2\times a\times 18.25

a=-5.76m/sec^2

We know that acceleration is given by

a=\mu g

5.76=\mu\times  9.81

\mu =0.587

So coefficient of friction will be 0.587

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A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

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2 years ago
How much work is done if a force of 20 N moves an object a distance of 6 m?​
Nesterboy [21]

Explanation:

W=F×s

w=20N×6m=120J

W=120J

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Michael's house is 5.0 km away from his school. How long would it take him to go ti school, riding a bus, if its velocity is 25
Anvisha [2.4K]

Answer:

12 mins

Explanation:

The distance covered is 5 km, divide this by 25 to get the fraction of an hour it takes. Doing this you get .2, times this by 60 min (1 hour) to get how many mins it takes

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