363326
Which one of the following statements is incorrect?
1 Frictional force opposes the relative motion
2 Coefficient of sliding friction has dimensions of length
3 Limiting value of static friction is directly proportional to normal reaction
4 Rolling friction is smaller than sliding friction.
Explanation:
Conceptual Question
NEET - 2018
PHXI05:LAWS OF MOTION
363327
A particle is projected with a speed \({v_0} = \sqrt {gR} \). The coefficient of friction between the particle and the hemispherical plane is \(\mu = 0.5\). Then, the acceleration of the particle is
1 \(g\)
2 \(\frac{{\sqrt 5 \,g}}{2}\)
3 \(\sqrt 2 \,g\)
4 \({\rm{None }}\,{\rm{of }}\,{\rm{these}}\)
Explanation:
When the particle slide on the surface then the friction will be kinetic in nature. This kinetic friction will provide tangential acceleration to the particle. Tangential acceleration of the particle is \({a_t} = \frac{{\mu mg}}{m} = \mu g = \frac{g}{2}\) Radial acceleration of the particle is \({a_n} = \frac{{{v^2}}}{R} = \frac{{gR}}{R} = g\) Net acceleration of the particle, \({a_{net}} = \sqrt {a_t^2 + a_n^2} = \sqrt {\frac{{{g^2}}}{4} + {g^2}} = \frac{{\sqrt 5 g}}{2}\)
PHXI05:LAWS OF MOTION
363328
A 100 \(N\) force acts horizontally on a block of mass 10 \(kg\) placed on a horizontal rough table of coefficient of friction \(\mu = 0.5\). If \(g\) at the place is \(10\,m{s^{ - 2}}\), the acceleration of the block is
363329
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 \(60\;N\) is applied on the body. The resulting acceleration of the body is about
1 \(6\;m{s^{ - 2}}\)
2 zero
3 \(1\;m{s^{ - 2}}\)
4 \(5\;m{s^{ - 2}}\)
Explanation:
Normal reaction \(R=m g\). We can take \(g = 10\,m{\rm{/}}{s^2}\) \({\mu _k} = 0.5\) \(f = {\mu _k}R = (0.5)(10)(10) = 50\,N\) \( \Rightarrow a = \frac{{F - f}}{m} = \frac{{60 - 50}}{{10}} = 1\,m{\rm{/}}{s^2}\) Correct option is (3).
363326
Which one of the following statements is incorrect?
1 Frictional force opposes the relative motion
2 Coefficient of sliding friction has dimensions of length
3 Limiting value of static friction is directly proportional to normal reaction
4 Rolling friction is smaller than sliding friction.
Explanation:
Conceptual Question
NEET - 2018
PHXI05:LAWS OF MOTION
363327
A particle is projected with a speed \({v_0} = \sqrt {gR} \). The coefficient of friction between the particle and the hemispherical plane is \(\mu = 0.5\). Then, the acceleration of the particle is
1 \(g\)
2 \(\frac{{\sqrt 5 \,g}}{2}\)
3 \(\sqrt 2 \,g\)
4 \({\rm{None }}\,{\rm{of }}\,{\rm{these}}\)
Explanation:
When the particle slide on the surface then the friction will be kinetic in nature. This kinetic friction will provide tangential acceleration to the particle. Tangential acceleration of the particle is \({a_t} = \frac{{\mu mg}}{m} = \mu g = \frac{g}{2}\) Radial acceleration of the particle is \({a_n} = \frac{{{v^2}}}{R} = \frac{{gR}}{R} = g\) Net acceleration of the particle, \({a_{net}} = \sqrt {a_t^2 + a_n^2} = \sqrt {\frac{{{g^2}}}{4} + {g^2}} = \frac{{\sqrt 5 g}}{2}\)
PHXI05:LAWS OF MOTION
363328
A 100 \(N\) force acts horizontally on a block of mass 10 \(kg\) placed on a horizontal rough table of coefficient of friction \(\mu = 0.5\). If \(g\) at the place is \(10\,m{s^{ - 2}}\), the acceleration of the block is
363329
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 \(60\;N\) is applied on the body. The resulting acceleration of the body is about
1 \(6\;m{s^{ - 2}}\)
2 zero
3 \(1\;m{s^{ - 2}}\)
4 \(5\;m{s^{ - 2}}\)
Explanation:
Normal reaction \(R=m g\). We can take \(g = 10\,m{\rm{/}}{s^2}\) \({\mu _k} = 0.5\) \(f = {\mu _k}R = (0.5)(10)(10) = 50\,N\) \( \Rightarrow a = \frac{{F - f}}{m} = \frac{{60 - 50}}{{10}} = 1\,m{\rm{/}}{s^2}\) Correct option is (3).
363326
Which one of the following statements is incorrect?
1 Frictional force opposes the relative motion
2 Coefficient of sliding friction has dimensions of length
3 Limiting value of static friction is directly proportional to normal reaction
4 Rolling friction is smaller than sliding friction.
Explanation:
Conceptual Question
NEET - 2018
PHXI05:LAWS OF MOTION
363327
A particle is projected with a speed \({v_0} = \sqrt {gR} \). The coefficient of friction between the particle and the hemispherical plane is \(\mu = 0.5\). Then, the acceleration of the particle is
1 \(g\)
2 \(\frac{{\sqrt 5 \,g}}{2}\)
3 \(\sqrt 2 \,g\)
4 \({\rm{None }}\,{\rm{of }}\,{\rm{these}}\)
Explanation:
When the particle slide on the surface then the friction will be kinetic in nature. This kinetic friction will provide tangential acceleration to the particle. Tangential acceleration of the particle is \({a_t} = \frac{{\mu mg}}{m} = \mu g = \frac{g}{2}\) Radial acceleration of the particle is \({a_n} = \frac{{{v^2}}}{R} = \frac{{gR}}{R} = g\) Net acceleration of the particle, \({a_{net}} = \sqrt {a_t^2 + a_n^2} = \sqrt {\frac{{{g^2}}}{4} + {g^2}} = \frac{{\sqrt 5 g}}{2}\)
PHXI05:LAWS OF MOTION
363328
A 100 \(N\) force acts horizontally on a block of mass 10 \(kg\) placed on a horizontal rough table of coefficient of friction \(\mu = 0.5\). If \(g\) at the place is \(10\,m{s^{ - 2}}\), the acceleration of the block is
363329
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 \(60\;N\) is applied on the body. The resulting acceleration of the body is about
1 \(6\;m{s^{ - 2}}\)
2 zero
3 \(1\;m{s^{ - 2}}\)
4 \(5\;m{s^{ - 2}}\)
Explanation:
Normal reaction \(R=m g\). We can take \(g = 10\,m{\rm{/}}{s^2}\) \({\mu _k} = 0.5\) \(f = {\mu _k}R = (0.5)(10)(10) = 50\,N\) \( \Rightarrow a = \frac{{F - f}}{m} = \frac{{60 - 50}}{{10}} = 1\,m{\rm{/}}{s^2}\) Correct option is (3).
363326
Which one of the following statements is incorrect?
1 Frictional force opposes the relative motion
2 Coefficient of sliding friction has dimensions of length
3 Limiting value of static friction is directly proportional to normal reaction
4 Rolling friction is smaller than sliding friction.
Explanation:
Conceptual Question
NEET - 2018
PHXI05:LAWS OF MOTION
363327
A particle is projected with a speed \({v_0} = \sqrt {gR} \). The coefficient of friction between the particle and the hemispherical plane is \(\mu = 0.5\). Then, the acceleration of the particle is
1 \(g\)
2 \(\frac{{\sqrt 5 \,g}}{2}\)
3 \(\sqrt 2 \,g\)
4 \({\rm{None }}\,{\rm{of }}\,{\rm{these}}\)
Explanation:
When the particle slide on the surface then the friction will be kinetic in nature. This kinetic friction will provide tangential acceleration to the particle. Tangential acceleration of the particle is \({a_t} = \frac{{\mu mg}}{m} = \mu g = \frac{g}{2}\) Radial acceleration of the particle is \({a_n} = \frac{{{v^2}}}{R} = \frac{{gR}}{R} = g\) Net acceleration of the particle, \({a_{net}} = \sqrt {a_t^2 + a_n^2} = \sqrt {\frac{{{g^2}}}{4} + {g^2}} = \frac{{\sqrt 5 g}}{2}\)
PHXI05:LAWS OF MOTION
363328
A 100 \(N\) force acts horizontally on a block of mass 10 \(kg\) placed on a horizontal rough table of coefficient of friction \(\mu = 0.5\). If \(g\) at the place is \(10\,m{s^{ - 2}}\), the acceleration of the block is
363329
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 \(60\;N\) is applied on the body. The resulting acceleration of the body is about
1 \(6\;m{s^{ - 2}}\)
2 zero
3 \(1\;m{s^{ - 2}}\)
4 \(5\;m{s^{ - 2}}\)
Explanation:
Normal reaction \(R=m g\). We can take \(g = 10\,m{\rm{/}}{s^2}\) \({\mu _k} = 0.5\) \(f = {\mu _k}R = (0.5)(10)(10) = 50\,N\) \( \Rightarrow a = \frac{{F - f}}{m} = \frac{{60 - 50}}{{10}} = 1\,m{\rm{/}}{s^2}\) Correct option is (3).
