KCET · Physics · Gravitation
A person is driving a vehicle at uniform speed of \( 5 \mathrm{~ms}^{-1} \) on a level curved track of radius \( 5 \mathrm{~m} \).
The coefficient of static friction between tyres and road is \( 0.1 \). Will the person slip while taking
the turn with the same speed? Take \( g=10 \mathrm{~ms}^{-2} \).
Choose the correct statement.
- A A person will slip if \( v^{2}=5 \mathrm{~ms}^{-1} \)
- B A person will slip if \( v^{2}>5 m s^{-1} \)
- C A person will slip if \( v^{2} < 5 m s^{-1} \)
- D A person will slip if \( v^{2}>10 \mathrm{~ms}^{-1} \)
Answer & Solution
Correct Answer
(B) A person will slip if \( v^{2}>5 m s^{-1} \)
Step-by-step Solution
Detailed explanation
Given, Speed of vehicle \(=5 \mathrm{~ms}^{-1}\), radius of track \(=5 \mathrm{~m}\), coefficient of static friction \(=0.1, \mathrm{~g}=10 \mathrm{~m} \mathrm{~s}-1\) maximum
speed without slipping in given as
\(\mathrm{v}_{\max }=\sqrt{\mu_{s} \times r \times g}\)
Substituting the values, we get
\(v_{\max }=\sqrt{0.1 \times 5 \times 10}=\sqrt{5} \Rightarrow v_{\max }^{2}=5\)
Thus, the person will slip while taking the turn with the speed greater than \(5 \mathrm{~ms}^{-1}\)
speed without slipping in given as
\(\mathrm{v}_{\max }=\sqrt{\mu_{s} \times r \times g}\)
Substituting the values, we get
\(v_{\max }=\sqrt{0.1 \times 5 \times 10}=\sqrt{5} \Rightarrow v_{\max }^{2}=5\)
Thus, the person will slip while taking the turn with the speed greater than \(5 \mathrm{~ms}^{-1}\)
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