Respuesta :
Answer:
speed on planet is 0.5 times of speed on a earth
Explanation:
Given data:
coeeffcient of kinetic friction[tex]= \mu+k =0.8[/tex]
friction force is given as
[tex]f_k = \mu_k N[/tex]n -normal force = mg
for max safe speed
fricional force = centripetal force
[tex]\mu_k N = \frac{mV^2}{R}[/tex]
[tex]\mu_k m g = \frac{mV^2}{R}[/tex]
[tex]v^2 = \mu_k g R[/tex]
[tex]v^2 = 0.8\times 2.45 \times 50[/tex]v^2 = 98
v = 9.899 m/sec
therefore maximum speed is [tex][v max} = 9.899 m/sec[/tex]
[tex]\frac{planet\ speed}{ earth\ speed} = \sqrt { \frac{ \mu_k g_p R}{ \mu_k g_e R}}[/tex]
[tex]\frac{V_p}{V_e} = \sqrt{\frac{g_p}{g_e}}[/tex]
solving for planet speed
[tex]v_p = \sqrt{\frac{2.45}{9.8}} v_e[/tex]
[tex]v_p = 0.5 v_e[/tex]
speed on planet is 0.5 times of speed on a earth
Answer:
b. 0.500
Explanation:
Given:
- coefficient of static friction, [tex]\mu_s=0.95[/tex]
- coefficient of kinetic friction, [tex]\mu_k=0.8[/tex]
- radius of circular track, [tex]r=50\ m[/tex]
- gravity on the planet, [tex]g=2.45\ m.s^{-2}[/tex]
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.
[tex]f=F_c[/tex]
[tex]\mu_s.R=m.\frac{v^2}{r}[/tex]
[tex]\mu_s.m.g=m.\frac{v^2}{r}[/tex]
[tex]\mu_s.g=\frac{v^2}{r}[/tex] ..........................................(1)
[tex]0.95\times 2.45=\frac{v^2}{50}[/tex]
[tex]v=10.7877\ m.s^{-2}[/tex]
Now putting values for the earth in eq. (1)
[tex]0.95\times 9.8=\frac{v_e^2}{50}[/tex]
[tex]v_e=21.5754\ m.s^{-1}[/tex]
Hence:
[tex]\frac{v}{v_e} =\frac{10.7877}{21.5754} =\frac{1}{2}[/tex]
