372141
A force of \(49 \mathrm{~N}\) is just able to move a block of wood weighing \(10 \mathrm{~kg}\) on a rough horizontal surface. Its coefficient of friction is
1 1
2 0.7
3 0.5
4 zero
Explanation:
C Given, \(\mathrm{F}=49 \mathrm{~N}, \mathrm{~m}=10 \mathrm{~kg}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The moment of block is just to move - Applied force \(=\) friction force \(\mathrm{F} =\mathrm{f}=\mu \mathrm{N}\) \(\mu =\frac{\mathrm{F}}{\mathrm{N}}=\frac{\mathrm{F}}{\mathrm{mg}} \quad(\therefore \mathrm{N}=\mathrm{mg})\) \(=\frac{49}{10 \times 10} \quad=0.49 \simeq 0.5\)
UP CPMT-2005
LAWS OF MOTION (ADDITIONAL)
372142
A car is moving at a speed of \(60 \mathrm{~km} / \mathrm{h}\) traversing a circular road track of radius \(60 \mathrm{~m}\). The minimum coefficient of friction to prevent the skidding of the car is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(25 / 54\)
2 \(21 / 54\)
3 \(15 / 44\)
4 \(21 / 44\)
Explanation:
A Given, \(\mathrm{v}=60 \mathrm{~km} / \mathrm{hr}=60 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=50 / 3 \mathrm{~m} / \mathrm{s}, \mathrm{r}=60 \mathrm{~m}\), \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The only force that can create centripetal acceleration while the car is moving on a flat, circular track and prevent it from, skidding is frictional force Since, Centripetal force \(=\) friction force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{v}^{2}}{\mathrm{rg}}\) \(\mu=\frac{(50 / 3)^{2}}{60 \times 10}\) \(\mu=25 / 54\)
TS EAMCET(Medical)-2015
LAWS OF MOTION (ADDITIONAL)
372143
Assuming that the coefficient of friction between the road and the tyre of a car is 0.4 , the maximum speed of the car on a turn of radius \(100 \mathrm{~m}\) on a level road will be:
1 \(10 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(30 \mathrm{~m} / \mathrm{s}\)
4 \(40 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, \(\mathrm{r}=100 \mathrm{~m}, \mu=0.4, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know that, Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.4 \times 100 \times 10}\) \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
MP PET -2013
LAWS OF MOTION (ADDITIONAL)
372144
A motor car is moving on a straight horizontal road with a speed of \(20 \mathrm{~m} / \mathrm{s}\). The coefficient of friction between the tyres and the road is 0.4 . The minimum distance in which the car can come to stop is:
1 \(50 \mathrm{~m}\)
2 \(125 \mathrm{~m}\)
3 \(100 \mathrm{~m}\)
4 \(150 \mathrm{~m}\)
Explanation:
A Given, Initial velocity (u) \(=20 \mathrm{~m} / \mathrm{s}\) Final velocity \((v)=0\) Coefficient of friction \((\mu)=0.4\) So, Maximum retardation force \(=\mu \mathrm{N}\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu . \mathrm{g}\) Using third equation of motion- \(\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{as}\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}} \quad(\mathrm{v}=0)\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mu \mathrm{g}}\) \(\mathrm{s}=\frac{20 \times 20}{2 \times 0.4 \times 10}\) \(\mathrm{~s}=50 \mathrm{~m}\)
MP PET -2009
LAWS OF MOTION (ADDITIONAL)
372145
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(40 \mathrm{~m}\). If the coefficient of friction is 1.28 . The maximum velocity with which the car can be moved, is
372141
A force of \(49 \mathrm{~N}\) is just able to move a block of wood weighing \(10 \mathrm{~kg}\) on a rough horizontal surface. Its coefficient of friction is
1 1
2 0.7
3 0.5
4 zero
Explanation:
C Given, \(\mathrm{F}=49 \mathrm{~N}, \mathrm{~m}=10 \mathrm{~kg}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The moment of block is just to move - Applied force \(=\) friction force \(\mathrm{F} =\mathrm{f}=\mu \mathrm{N}\) \(\mu =\frac{\mathrm{F}}{\mathrm{N}}=\frac{\mathrm{F}}{\mathrm{mg}} \quad(\therefore \mathrm{N}=\mathrm{mg})\) \(=\frac{49}{10 \times 10} \quad=0.49 \simeq 0.5\)
UP CPMT-2005
LAWS OF MOTION (ADDITIONAL)
372142
A car is moving at a speed of \(60 \mathrm{~km} / \mathrm{h}\) traversing a circular road track of radius \(60 \mathrm{~m}\). The minimum coefficient of friction to prevent the skidding of the car is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(25 / 54\)
2 \(21 / 54\)
3 \(15 / 44\)
4 \(21 / 44\)
Explanation:
A Given, \(\mathrm{v}=60 \mathrm{~km} / \mathrm{hr}=60 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=50 / 3 \mathrm{~m} / \mathrm{s}, \mathrm{r}=60 \mathrm{~m}\), \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The only force that can create centripetal acceleration while the car is moving on a flat, circular track and prevent it from, skidding is frictional force Since, Centripetal force \(=\) friction force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{v}^{2}}{\mathrm{rg}}\) \(\mu=\frac{(50 / 3)^{2}}{60 \times 10}\) \(\mu=25 / 54\)
TS EAMCET(Medical)-2015
LAWS OF MOTION (ADDITIONAL)
372143
Assuming that the coefficient of friction between the road and the tyre of a car is 0.4 , the maximum speed of the car on a turn of radius \(100 \mathrm{~m}\) on a level road will be:
1 \(10 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(30 \mathrm{~m} / \mathrm{s}\)
4 \(40 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, \(\mathrm{r}=100 \mathrm{~m}, \mu=0.4, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know that, Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.4 \times 100 \times 10}\) \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
MP PET -2013
LAWS OF MOTION (ADDITIONAL)
372144
A motor car is moving on a straight horizontal road with a speed of \(20 \mathrm{~m} / \mathrm{s}\). The coefficient of friction between the tyres and the road is 0.4 . The minimum distance in which the car can come to stop is:
1 \(50 \mathrm{~m}\)
2 \(125 \mathrm{~m}\)
3 \(100 \mathrm{~m}\)
4 \(150 \mathrm{~m}\)
Explanation:
A Given, Initial velocity (u) \(=20 \mathrm{~m} / \mathrm{s}\) Final velocity \((v)=0\) Coefficient of friction \((\mu)=0.4\) So, Maximum retardation force \(=\mu \mathrm{N}\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu . \mathrm{g}\) Using third equation of motion- \(\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{as}\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}} \quad(\mathrm{v}=0)\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mu \mathrm{g}}\) \(\mathrm{s}=\frac{20 \times 20}{2 \times 0.4 \times 10}\) \(\mathrm{~s}=50 \mathrm{~m}\)
MP PET -2009
LAWS OF MOTION (ADDITIONAL)
372145
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(40 \mathrm{~m}\). If the coefficient of friction is 1.28 . The maximum velocity with which the car can be moved, is
372141
A force of \(49 \mathrm{~N}\) is just able to move a block of wood weighing \(10 \mathrm{~kg}\) on a rough horizontal surface. Its coefficient of friction is
1 1
2 0.7
3 0.5
4 zero
Explanation:
C Given, \(\mathrm{F}=49 \mathrm{~N}, \mathrm{~m}=10 \mathrm{~kg}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The moment of block is just to move - Applied force \(=\) friction force \(\mathrm{F} =\mathrm{f}=\mu \mathrm{N}\) \(\mu =\frac{\mathrm{F}}{\mathrm{N}}=\frac{\mathrm{F}}{\mathrm{mg}} \quad(\therefore \mathrm{N}=\mathrm{mg})\) \(=\frac{49}{10 \times 10} \quad=0.49 \simeq 0.5\)
UP CPMT-2005
LAWS OF MOTION (ADDITIONAL)
372142
A car is moving at a speed of \(60 \mathrm{~km} / \mathrm{h}\) traversing a circular road track of radius \(60 \mathrm{~m}\). The minimum coefficient of friction to prevent the skidding of the car is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(25 / 54\)
2 \(21 / 54\)
3 \(15 / 44\)
4 \(21 / 44\)
Explanation:
A Given, \(\mathrm{v}=60 \mathrm{~km} / \mathrm{hr}=60 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=50 / 3 \mathrm{~m} / \mathrm{s}, \mathrm{r}=60 \mathrm{~m}\), \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The only force that can create centripetal acceleration while the car is moving on a flat, circular track and prevent it from, skidding is frictional force Since, Centripetal force \(=\) friction force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{v}^{2}}{\mathrm{rg}}\) \(\mu=\frac{(50 / 3)^{2}}{60 \times 10}\) \(\mu=25 / 54\)
TS EAMCET(Medical)-2015
LAWS OF MOTION (ADDITIONAL)
372143
Assuming that the coefficient of friction between the road and the tyre of a car is 0.4 , the maximum speed of the car on a turn of radius \(100 \mathrm{~m}\) on a level road will be:
1 \(10 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(30 \mathrm{~m} / \mathrm{s}\)
4 \(40 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, \(\mathrm{r}=100 \mathrm{~m}, \mu=0.4, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know that, Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.4 \times 100 \times 10}\) \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
MP PET -2013
LAWS OF MOTION (ADDITIONAL)
372144
A motor car is moving on a straight horizontal road with a speed of \(20 \mathrm{~m} / \mathrm{s}\). The coefficient of friction between the tyres and the road is 0.4 . The minimum distance in which the car can come to stop is:
1 \(50 \mathrm{~m}\)
2 \(125 \mathrm{~m}\)
3 \(100 \mathrm{~m}\)
4 \(150 \mathrm{~m}\)
Explanation:
A Given, Initial velocity (u) \(=20 \mathrm{~m} / \mathrm{s}\) Final velocity \((v)=0\) Coefficient of friction \((\mu)=0.4\) So, Maximum retardation force \(=\mu \mathrm{N}\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu . \mathrm{g}\) Using third equation of motion- \(\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{as}\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}} \quad(\mathrm{v}=0)\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mu \mathrm{g}}\) \(\mathrm{s}=\frac{20 \times 20}{2 \times 0.4 \times 10}\) \(\mathrm{~s}=50 \mathrm{~m}\)
MP PET -2009
LAWS OF MOTION (ADDITIONAL)
372145
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(40 \mathrm{~m}\). If the coefficient of friction is 1.28 . The maximum velocity with which the car can be moved, is
372141
A force of \(49 \mathrm{~N}\) is just able to move a block of wood weighing \(10 \mathrm{~kg}\) on a rough horizontal surface. Its coefficient of friction is
1 1
2 0.7
3 0.5
4 zero
Explanation:
C Given, \(\mathrm{F}=49 \mathrm{~N}, \mathrm{~m}=10 \mathrm{~kg}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The moment of block is just to move - Applied force \(=\) friction force \(\mathrm{F} =\mathrm{f}=\mu \mathrm{N}\) \(\mu =\frac{\mathrm{F}}{\mathrm{N}}=\frac{\mathrm{F}}{\mathrm{mg}} \quad(\therefore \mathrm{N}=\mathrm{mg})\) \(=\frac{49}{10 \times 10} \quad=0.49 \simeq 0.5\)
UP CPMT-2005
LAWS OF MOTION (ADDITIONAL)
372142
A car is moving at a speed of \(60 \mathrm{~km} / \mathrm{h}\) traversing a circular road track of radius \(60 \mathrm{~m}\). The minimum coefficient of friction to prevent the skidding of the car is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(25 / 54\)
2 \(21 / 54\)
3 \(15 / 44\)
4 \(21 / 44\)
Explanation:
A Given, \(\mathrm{v}=60 \mathrm{~km} / \mathrm{hr}=60 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=50 / 3 \mathrm{~m} / \mathrm{s}, \mathrm{r}=60 \mathrm{~m}\), \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The only force that can create centripetal acceleration while the car is moving on a flat, circular track and prevent it from, skidding is frictional force Since, Centripetal force \(=\) friction force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{v}^{2}}{\mathrm{rg}}\) \(\mu=\frac{(50 / 3)^{2}}{60 \times 10}\) \(\mu=25 / 54\)
TS EAMCET(Medical)-2015
LAWS OF MOTION (ADDITIONAL)
372143
Assuming that the coefficient of friction between the road and the tyre of a car is 0.4 , the maximum speed of the car on a turn of radius \(100 \mathrm{~m}\) on a level road will be:
1 \(10 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(30 \mathrm{~m} / \mathrm{s}\)
4 \(40 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, \(\mathrm{r}=100 \mathrm{~m}, \mu=0.4, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know that, Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.4 \times 100 \times 10}\) \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
MP PET -2013
LAWS OF MOTION (ADDITIONAL)
372144
A motor car is moving on a straight horizontal road with a speed of \(20 \mathrm{~m} / \mathrm{s}\). The coefficient of friction between the tyres and the road is 0.4 . The minimum distance in which the car can come to stop is:
1 \(50 \mathrm{~m}\)
2 \(125 \mathrm{~m}\)
3 \(100 \mathrm{~m}\)
4 \(150 \mathrm{~m}\)
Explanation:
A Given, Initial velocity (u) \(=20 \mathrm{~m} / \mathrm{s}\) Final velocity \((v)=0\) Coefficient of friction \((\mu)=0.4\) So, Maximum retardation force \(=\mu \mathrm{N}\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu . \mathrm{g}\) Using third equation of motion- \(\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{as}\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}} \quad(\mathrm{v}=0)\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mu \mathrm{g}}\) \(\mathrm{s}=\frac{20 \times 20}{2 \times 0.4 \times 10}\) \(\mathrm{~s}=50 \mathrm{~m}\)
MP PET -2009
LAWS OF MOTION (ADDITIONAL)
372145
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(40 \mathrm{~m}\). If the coefficient of friction is 1.28 . The maximum velocity with which the car can be moved, is
372141
A force of \(49 \mathrm{~N}\) is just able to move a block of wood weighing \(10 \mathrm{~kg}\) on a rough horizontal surface. Its coefficient of friction is
1 1
2 0.7
3 0.5
4 zero
Explanation:
C Given, \(\mathrm{F}=49 \mathrm{~N}, \mathrm{~m}=10 \mathrm{~kg}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The moment of block is just to move - Applied force \(=\) friction force \(\mathrm{F} =\mathrm{f}=\mu \mathrm{N}\) \(\mu =\frac{\mathrm{F}}{\mathrm{N}}=\frac{\mathrm{F}}{\mathrm{mg}} \quad(\therefore \mathrm{N}=\mathrm{mg})\) \(=\frac{49}{10 \times 10} \quad=0.49 \simeq 0.5\)
UP CPMT-2005
LAWS OF MOTION (ADDITIONAL)
372142
A car is moving at a speed of \(60 \mathrm{~km} / \mathrm{h}\) traversing a circular road track of radius \(60 \mathrm{~m}\). The minimum coefficient of friction to prevent the skidding of the car is \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\)
1 \(25 / 54\)
2 \(21 / 54\)
3 \(15 / 44\)
4 \(21 / 44\)
Explanation:
A Given, \(\mathrm{v}=60 \mathrm{~km} / \mathrm{hr}=60 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=50 / 3 \mathrm{~m} / \mathrm{s}, \mathrm{r}=60 \mathrm{~m}\), \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) The only force that can create centripetal acceleration while the car is moving on a flat, circular track and prevent it from, skidding is frictional force Since, Centripetal force \(=\) friction force \(\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mu \mathrm{mg}\) \(\mu=\frac{\mathrm{v}^{2}}{\mathrm{rg}}\) \(\mu=\frac{(50 / 3)^{2}}{60 \times 10}\) \(\mu=25 / 54\)
TS EAMCET(Medical)-2015
LAWS OF MOTION (ADDITIONAL)
372143
Assuming that the coefficient of friction between the road and the tyre of a car is 0.4 , the maximum speed of the car on a turn of radius \(100 \mathrm{~m}\) on a level road will be:
1 \(10 \mathrm{~m} / \mathrm{s}\)
2 \(20 \mathrm{~m} / \mathrm{s}\)
3 \(30 \mathrm{~m} / \mathrm{s}\)
4 \(40 \mathrm{~m} / \mathrm{s}\)
Explanation:
B Given, \(\mathrm{r}=100 \mathrm{~m}, \mu=0.4, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}\) We know that, Frictional force \(=\) Centripetal force \(\mu \mathrm{mg}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\) \(\mathrm{v}^{2}=\mu \mathrm{rg}\) \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}\) \(\mathrm{v}=\sqrt{0.4 \times 100 \times 10}\) \(\mathrm{v}=20 \mathrm{~m} / \mathrm{s}\)
MP PET -2013
LAWS OF MOTION (ADDITIONAL)
372144
A motor car is moving on a straight horizontal road with a speed of \(20 \mathrm{~m} / \mathrm{s}\). The coefficient of friction between the tyres and the road is 0.4 . The minimum distance in which the car can come to stop is:
1 \(50 \mathrm{~m}\)
2 \(125 \mathrm{~m}\)
3 \(100 \mathrm{~m}\)
4 \(150 \mathrm{~m}\)
Explanation:
A Given, Initial velocity (u) \(=20 \mathrm{~m} / \mathrm{s}\) Final velocity \((v)=0\) Coefficient of friction \((\mu)=0.4\) So, Maximum retardation force \(=\mu \mathrm{N}\) \(\mathrm{ma}=\mu \mathrm{mg}\) \(\mathrm{a}=\mu . \mathrm{g}\) Using third equation of motion- \(\mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{as}\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mathrm{a}} \quad(\mathrm{v}=0)\) \(\mathrm{s}=\frac{\mathrm{u}^{2}}{2 \mu \mathrm{g}}\) \(\mathrm{s}=\frac{20 \times 20}{2 \times 0.4 \times 10}\) \(\mathrm{~s}=50 \mathrm{~m}\)
MP PET -2009
LAWS OF MOTION (ADDITIONAL)
372145
A car of mass \(1000 \mathrm{~kg}\) moves on a circular track of radius \(40 \mathrm{~m}\). If the coefficient of friction is 1.28 . The maximum velocity with which the car can be moved, is
