372090
When the two surfaces are coated with the lubricant, then they will:
1 slide upon each other
2 stick to each other
3 roll upon each other
4 none of these
Explanation:
A Friction prevents relative motion between two surface. If two surface are coated with lubricant then friction will be reduced so they can slide over each other if one is pushed on the other.
AIIMS-2001
LAWS OF MOTION (ADDITIONAL)
372091
If the coefficient of static friction between the tyres and road is 0.5 , what is the shortest distance in which an automobile can be stopped when travelling at \(72 \mathrm{~km} / \mathrm{h}\) ?
1 \(50 \mathrm{~m}\)
2 \(60 \mathrm{~m}\)
3 \(40.8 \mathrm{~m}\)
4 \(80.16 \mathrm{~m}\)
Explanation:
C Given, \(\mathrm{v} =0\) \(\mathrm{u} =72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}\) \(\mathrm{g} =9.8 \mathrm{~m} / \mathrm{s}^{2}\) Here, frictional force working between road and tyres which is retards the motion of automobile. The static friction \(\left(f_{s}\right)\) working between tyres and road, so frictional force cause the retardation in velocity of automobile. Hence, \(\mathrm{F}=\mathrm{f}_{\mathrm{s}} =\mu \mathrm{R}\) \(=\mu . \mathrm{mg}\) Where \(m\) is the mass of automobile We know that, \(\mathrm{F}=\mathrm{ma}\) \(\mu \mathrm{mg}=\mathrm{ma}\) \(\mathrm{a}=\mu \mathrm{g}\) \(\mathrm{a}=0.5 \mathrm{~g}\) Retardation is \(0.5 \mathrm{~g}\) Let automobile stops at a distance \(\mathrm{x}\), then from third equation of motion- \(v^{2} =u^{2}-2 \text { as }\) \(0^{2} =(20)^{2}-2 \times(0.5 \times 9.8) x\) \(x =\frac{20 \times 20}{2 \times 0.5 \times 9.8}=40.8 \mathrm{~m}\) \(\therefore \quad 0^{2}=(20)^{2}-2 \times(0.5 \times 9.8) \mathrm{x}\)
BCECE-2007
LAWS OF MOTION (ADDITIONAL)
372092
In the given figure the pulley is assumed massless and frictionless. If the friction force on the object of mass \(m\) is \(f\), then its acceleration in terms of the force \(F\) will be eaual to :
B As the pulley is mass-less and frictionless, net force on pulley will be zero. According to question - \(\frac{\mathrm{F}}{2}-\mathrm{f}=\mathrm{ma}\) \(\mathrm{a}=\frac{\left(\frac{\mathrm{F}}{2}-\mathrm{f}\right)}{\mathrm{m}}\)
BCECE-2006
LAWS OF MOTION (ADDITIONAL)
372093
A body is moving along a rough horizontal surface with an initial velocity \(6 \mathrm{~m} / \mathrm{s}\). If the body comes to rest after travelling a distance 9 \(\mathrm{m}\), then the coefficient of sliding friction will be:
1 0.4
2 0.2
3 0.6
4 0.8
Explanation:
B Given, Initial velocity \((u)=6 \mathrm{~m} / \mathrm{s}, \mathrm{s}=9 \mathrm{~m}, \mu_{\mathrm{s}}=\) ? We know that, \(\mathrm{F} =\mu_{\mathrm{s}} \cdot \mathrm{N}\) \(=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) If acceleration of body is a Then, \(\mathrm{F}=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) \(\mathrm{m} \cdot \mathrm{a}=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) \(\mu_{\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{g}}\) Where, \(\mu_{\mathrm{s}}\) is sliding friction. Apply equation of motion - \(v^{2} =u^{2}+2 a s\) \(0 =(6)^{2}+2 a \times 9\) \(-36 =18 a\) \(\mathrm{a}=-2 \mathrm{~m} / \mathrm{s}^{2}\) (-ve sign indicate body is retarding) Hence, coefficient of sliding friction is - \(\mu_{\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{g}}=\frac{2}{10}=0.2\)
372090
When the two surfaces are coated with the lubricant, then they will:
1 slide upon each other
2 stick to each other
3 roll upon each other
4 none of these
Explanation:
A Friction prevents relative motion between two surface. If two surface are coated with lubricant then friction will be reduced so they can slide over each other if one is pushed on the other.
AIIMS-2001
LAWS OF MOTION (ADDITIONAL)
372091
If the coefficient of static friction between the tyres and road is 0.5 , what is the shortest distance in which an automobile can be stopped when travelling at \(72 \mathrm{~km} / \mathrm{h}\) ?
1 \(50 \mathrm{~m}\)
2 \(60 \mathrm{~m}\)
3 \(40.8 \mathrm{~m}\)
4 \(80.16 \mathrm{~m}\)
Explanation:
C Given, \(\mathrm{v} =0\) \(\mathrm{u} =72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}\) \(\mathrm{g} =9.8 \mathrm{~m} / \mathrm{s}^{2}\) Here, frictional force working between road and tyres which is retards the motion of automobile. The static friction \(\left(f_{s}\right)\) working between tyres and road, so frictional force cause the retardation in velocity of automobile. Hence, \(\mathrm{F}=\mathrm{f}_{\mathrm{s}} =\mu \mathrm{R}\) \(=\mu . \mathrm{mg}\) Where \(m\) is the mass of automobile We know that, \(\mathrm{F}=\mathrm{ma}\) \(\mu \mathrm{mg}=\mathrm{ma}\) \(\mathrm{a}=\mu \mathrm{g}\) \(\mathrm{a}=0.5 \mathrm{~g}\) Retardation is \(0.5 \mathrm{~g}\) Let automobile stops at a distance \(\mathrm{x}\), then from third equation of motion- \(v^{2} =u^{2}-2 \text { as }\) \(0^{2} =(20)^{2}-2 \times(0.5 \times 9.8) x\) \(x =\frac{20 \times 20}{2 \times 0.5 \times 9.8}=40.8 \mathrm{~m}\) \(\therefore \quad 0^{2}=(20)^{2}-2 \times(0.5 \times 9.8) \mathrm{x}\)
BCECE-2007
LAWS OF MOTION (ADDITIONAL)
372092
In the given figure the pulley is assumed massless and frictionless. If the friction force on the object of mass \(m\) is \(f\), then its acceleration in terms of the force \(F\) will be eaual to :
B As the pulley is mass-less and frictionless, net force on pulley will be zero. According to question - \(\frac{\mathrm{F}}{2}-\mathrm{f}=\mathrm{ma}\) \(\mathrm{a}=\frac{\left(\frac{\mathrm{F}}{2}-\mathrm{f}\right)}{\mathrm{m}}\)
BCECE-2006
LAWS OF MOTION (ADDITIONAL)
372093
A body is moving along a rough horizontal surface with an initial velocity \(6 \mathrm{~m} / \mathrm{s}\). If the body comes to rest after travelling a distance 9 \(\mathrm{m}\), then the coefficient of sliding friction will be:
1 0.4
2 0.2
3 0.6
4 0.8
Explanation:
B Given, Initial velocity \((u)=6 \mathrm{~m} / \mathrm{s}, \mathrm{s}=9 \mathrm{~m}, \mu_{\mathrm{s}}=\) ? We know that, \(\mathrm{F} =\mu_{\mathrm{s}} \cdot \mathrm{N}\) \(=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) If acceleration of body is a Then, \(\mathrm{F}=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) \(\mathrm{m} \cdot \mathrm{a}=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) \(\mu_{\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{g}}\) Where, \(\mu_{\mathrm{s}}\) is sliding friction. Apply equation of motion - \(v^{2} =u^{2}+2 a s\) \(0 =(6)^{2}+2 a \times 9\) \(-36 =18 a\) \(\mathrm{a}=-2 \mathrm{~m} / \mathrm{s}^{2}\) (-ve sign indicate body is retarding) Hence, coefficient of sliding friction is - \(\mu_{\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{g}}=\frac{2}{10}=0.2\)
372090
When the two surfaces are coated with the lubricant, then they will:
1 slide upon each other
2 stick to each other
3 roll upon each other
4 none of these
Explanation:
A Friction prevents relative motion between two surface. If two surface are coated with lubricant then friction will be reduced so they can slide over each other if one is pushed on the other.
AIIMS-2001
LAWS OF MOTION (ADDITIONAL)
372091
If the coefficient of static friction between the tyres and road is 0.5 , what is the shortest distance in which an automobile can be stopped when travelling at \(72 \mathrm{~km} / \mathrm{h}\) ?
1 \(50 \mathrm{~m}\)
2 \(60 \mathrm{~m}\)
3 \(40.8 \mathrm{~m}\)
4 \(80.16 \mathrm{~m}\)
Explanation:
C Given, \(\mathrm{v} =0\) \(\mathrm{u} =72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}\) \(\mathrm{g} =9.8 \mathrm{~m} / \mathrm{s}^{2}\) Here, frictional force working between road and tyres which is retards the motion of automobile. The static friction \(\left(f_{s}\right)\) working between tyres and road, so frictional force cause the retardation in velocity of automobile. Hence, \(\mathrm{F}=\mathrm{f}_{\mathrm{s}} =\mu \mathrm{R}\) \(=\mu . \mathrm{mg}\) Where \(m\) is the mass of automobile We know that, \(\mathrm{F}=\mathrm{ma}\) \(\mu \mathrm{mg}=\mathrm{ma}\) \(\mathrm{a}=\mu \mathrm{g}\) \(\mathrm{a}=0.5 \mathrm{~g}\) Retardation is \(0.5 \mathrm{~g}\) Let automobile stops at a distance \(\mathrm{x}\), then from third equation of motion- \(v^{2} =u^{2}-2 \text { as }\) \(0^{2} =(20)^{2}-2 \times(0.5 \times 9.8) x\) \(x =\frac{20 \times 20}{2 \times 0.5 \times 9.8}=40.8 \mathrm{~m}\) \(\therefore \quad 0^{2}=(20)^{2}-2 \times(0.5 \times 9.8) \mathrm{x}\)
BCECE-2007
LAWS OF MOTION (ADDITIONAL)
372092
In the given figure the pulley is assumed massless and frictionless. If the friction force on the object of mass \(m\) is \(f\), then its acceleration in terms of the force \(F\) will be eaual to :
B As the pulley is mass-less and frictionless, net force on pulley will be zero. According to question - \(\frac{\mathrm{F}}{2}-\mathrm{f}=\mathrm{ma}\) \(\mathrm{a}=\frac{\left(\frac{\mathrm{F}}{2}-\mathrm{f}\right)}{\mathrm{m}}\)
BCECE-2006
LAWS OF MOTION (ADDITIONAL)
372093
A body is moving along a rough horizontal surface with an initial velocity \(6 \mathrm{~m} / \mathrm{s}\). If the body comes to rest after travelling a distance 9 \(\mathrm{m}\), then the coefficient of sliding friction will be:
1 0.4
2 0.2
3 0.6
4 0.8
Explanation:
B Given, Initial velocity \((u)=6 \mathrm{~m} / \mathrm{s}, \mathrm{s}=9 \mathrm{~m}, \mu_{\mathrm{s}}=\) ? We know that, \(\mathrm{F} =\mu_{\mathrm{s}} \cdot \mathrm{N}\) \(=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) If acceleration of body is a Then, \(\mathrm{F}=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) \(\mathrm{m} \cdot \mathrm{a}=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) \(\mu_{\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{g}}\) Where, \(\mu_{\mathrm{s}}\) is sliding friction. Apply equation of motion - \(v^{2} =u^{2}+2 a s\) \(0 =(6)^{2}+2 a \times 9\) \(-36 =18 a\) \(\mathrm{a}=-2 \mathrm{~m} / \mathrm{s}^{2}\) (-ve sign indicate body is retarding) Hence, coefficient of sliding friction is - \(\mu_{\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{g}}=\frac{2}{10}=0.2\)
372090
When the two surfaces are coated with the lubricant, then they will:
1 slide upon each other
2 stick to each other
3 roll upon each other
4 none of these
Explanation:
A Friction prevents relative motion between two surface. If two surface are coated with lubricant then friction will be reduced so they can slide over each other if one is pushed on the other.
AIIMS-2001
LAWS OF MOTION (ADDITIONAL)
372091
If the coefficient of static friction between the tyres and road is 0.5 , what is the shortest distance in which an automobile can be stopped when travelling at \(72 \mathrm{~km} / \mathrm{h}\) ?
1 \(50 \mathrm{~m}\)
2 \(60 \mathrm{~m}\)
3 \(40.8 \mathrm{~m}\)
4 \(80.16 \mathrm{~m}\)
Explanation:
C Given, \(\mathrm{v} =0\) \(\mathrm{u} =72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}\) \(\mathrm{g} =9.8 \mathrm{~m} / \mathrm{s}^{2}\) Here, frictional force working between road and tyres which is retards the motion of automobile. The static friction \(\left(f_{s}\right)\) working between tyres and road, so frictional force cause the retardation in velocity of automobile. Hence, \(\mathrm{F}=\mathrm{f}_{\mathrm{s}} =\mu \mathrm{R}\) \(=\mu . \mathrm{mg}\) Where \(m\) is the mass of automobile We know that, \(\mathrm{F}=\mathrm{ma}\) \(\mu \mathrm{mg}=\mathrm{ma}\) \(\mathrm{a}=\mu \mathrm{g}\) \(\mathrm{a}=0.5 \mathrm{~g}\) Retardation is \(0.5 \mathrm{~g}\) Let automobile stops at a distance \(\mathrm{x}\), then from third equation of motion- \(v^{2} =u^{2}-2 \text { as }\) \(0^{2} =(20)^{2}-2 \times(0.5 \times 9.8) x\) \(x =\frac{20 \times 20}{2 \times 0.5 \times 9.8}=40.8 \mathrm{~m}\) \(\therefore \quad 0^{2}=(20)^{2}-2 \times(0.5 \times 9.8) \mathrm{x}\)
BCECE-2007
LAWS OF MOTION (ADDITIONAL)
372092
In the given figure the pulley is assumed massless and frictionless. If the friction force on the object of mass \(m\) is \(f\), then its acceleration in terms of the force \(F\) will be eaual to :
B As the pulley is mass-less and frictionless, net force on pulley will be zero. According to question - \(\frac{\mathrm{F}}{2}-\mathrm{f}=\mathrm{ma}\) \(\mathrm{a}=\frac{\left(\frac{\mathrm{F}}{2}-\mathrm{f}\right)}{\mathrm{m}}\)
BCECE-2006
LAWS OF MOTION (ADDITIONAL)
372093
A body is moving along a rough horizontal surface with an initial velocity \(6 \mathrm{~m} / \mathrm{s}\). If the body comes to rest after travelling a distance 9 \(\mathrm{m}\), then the coefficient of sliding friction will be:
1 0.4
2 0.2
3 0.6
4 0.8
Explanation:
B Given, Initial velocity \((u)=6 \mathrm{~m} / \mathrm{s}, \mathrm{s}=9 \mathrm{~m}, \mu_{\mathrm{s}}=\) ? We know that, \(\mathrm{F} =\mu_{\mathrm{s}} \cdot \mathrm{N}\) \(=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) If acceleration of body is a Then, \(\mathrm{F}=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) \(\mathrm{m} \cdot \mathrm{a}=\mu_{\mathrm{s}} \cdot \mathrm{mg}\) \(\mu_{\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{g}}\) Where, \(\mu_{\mathrm{s}}\) is sliding friction. Apply equation of motion - \(v^{2} =u^{2}+2 a s\) \(0 =(6)^{2}+2 a \times 9\) \(-36 =18 a\) \(\mathrm{a}=-2 \mathrm{~m} / \mathrm{s}^{2}\) (-ve sign indicate body is retarding) Hence, coefficient of sliding friction is - \(\mu_{\mathrm{s}}=\frac{\mathrm{a}}{\mathrm{g}}=\frac{2}{10}=0.2\)
