KCET · Physics · Laws of Motion
A body of mass \(10 \mathrm{~kg}\) is kept on a horizontal surface. The coefficient of kinetic friction between the body and the surface is 0.5 . A horizontal force of \(60 \mathrm{~N}\) is applied on the body. The resulting acceleration of the body is about
- A \(1 \mathrm{~ms}^{-2}\)
- B \(5 \mathrm{~ms}^{-2}\)
- C \(6 \mathrm{~ms}^{-2}\)
- D zero
Answer & Solution
Correct Answer
(A) \(1 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Given, mass of the body, \(m=10 \mathrm{~kg}\)
Coefficient of kinetic friction, \(\mu_k=0.5\)
Force applied in horizontal direction, \(F=60 \mathrm{~N}\)
Let \(f\) be the frictional force on the body.
Thus, \(m a=F-\mu m g\)
\(=60-10 \times(10)(0.9)=60-50\)
\(\Rightarrow 10 \times a=10 \Rightarrow a=1 \mathrm{~m} / \mathrm{s}^2\)
Coefficient of kinetic friction, \(\mu_k=0.5\)
Force applied in horizontal direction, \(F=60 \mathrm{~N}\)
Let \(f\) be the frictional force on the body.
Thus, \(m a=F-\mu m g\)
\(=60-10 \times(10)(0.9)=60-50\)
\(\Rightarrow 10 \times a=10 \Rightarrow a=1 \mathrm{~m} / \mathrm{s}^2\)
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