Question

The coefficient of static friction for the tires of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The car is on a level circular track of 50.0 m radius on a planet where g = 2.45 m/s² compared to Earth's g=9.8m/s². The maximum safe speed on the track on the planet is ____ times as large as the maximumsafe speed on Earth.a. 0.250b. 0.500c. 1.00d. 2.00e. 4.00

Answer 1

b. 0.500

Step-by-step explanation:

Given:

• coefficient of static friction,
  `\mu_s=0.95`

• coefficient of kinetic friction,
  `\mu_k=0.8`

• radius of circular track,
  `r=50\ m`

• gravity on the planet,
  `g=2.45\ m.s^(-2)`

For the condition of safe speed the centripetal force must be equal to the static frictional force so that the wheels of the car do not slip in tangential direction or into the curvature of turn.

`f=F_c`

`\mu_s.R=m.(v^2)/(r)`

`\mu_s.m.g=m.(v^2)/(r)`

`\mu_s.g=(v^2)/(r)` ..........................................(1)

`0.95* 2.45=(v^2)/(50)`

`v=10.7877\ m.s^(-2)`

Now putting values for the earth in eq. (1)

`0.95* 9.8=(v_e^2)/(50)`

`v_e=21.5754\ m.s^(-1)`

Hence:

`(v)/(v_e) =(10.7877)/(21.5754) =(1)/(2)`

Answer 2

speed on planet is 0.5 times of speed on a earth

Step-by-step explanation:

Given data:

coeeffcient of kinetic friction
`= \mu+k  =0.8`

friction force is given as

`f_k = \mu_k N`n -normal force = mg

for max safe speed

fricional force = centripetal force

`\mu_k N = (mV^2)/(R)`

`\mu_k m g = (mV^2)/(R)`

`v^2 = \mu_k g R`

`v^2 = 0.8* 2.45 * 50`v^2 = 98

v = 9.899 m/sec

therefore maximum speed is
`[v max} = 9.899 m/sec`

`(planet\ speed)/( earth\ speed) = \sqrt { ( \mu_k g_p R)/( \mu_k g_e R)}`

`(V_p)/(V_e) = \sqrt{(g_p)/(g_e)}`

solving for planet speed

`v_p = \sqrt{(2.45)/(9.8)} v_e`

`v_p = 0.5 v_e`

speed on planet is 0.5 times of speed on a earth