(A) The work done by friction is given by $W_f = -\int \mu N \, ds$. For route $1$,the surface is convex,so the normal force $N_1 = mg \cos \theta - \

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(A) The work done by friction is given by $W_f = -\int \mu N \, ds$. For route $1$,the surface is convex,so the normal force $N_1 = mg \cos \theta - \frac{mv^2}{r}$. For route $2$,the surface is concave,so the normal force $N_2 = mg \cos \theta + \frac{mv^2}{r}$. Since $N_2 > N_1$,the frictional force $f = \mu N$ is greater on route $2$ than on route $1$. As the work done against friction is greater on route $2$,more mechanical energy is dissipated as heat on route $2$. Therefore,the final kinetic energy and speed at point $B$ will be greater for route $1$.

Class 11 Physics Newton's Laws of Motion and Friction Kinetic Friction and Motion on Rough Horizontal Surface

Medium

$A$ small body slips,subject to the force of friction,from point $A$ to point $B$ along two curved surfaces of equal radius,first along route $1$,then along route $2$. Friction does not depend on the speed and the coefficient of friction on both routes is the same. In which case will the body's speed at $B$ be greater?

A
speed is greater in case $1$
B
speed is greater in case $2$
C
speed is same in both cases
D
cannot be determined

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Class 11 Questions

Class 11

Physics Questions

Chapter

Newton's Laws of Motion and Friction

Subtopic

Kinetic Friction and Motion on Rough Horizontal Surface

The retarding acceleration of $7.35\, ms^{-2}$ due to frictional force stops a car of mass $400\, kg$ travelling on a road. The coefficient of friction between the tyre of the car and the road is

Medium

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$A$ block of mass $m$ slides along a floor while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in the figure. The coefficient of kinetic friction is $\mu_{K}$. Then,the block's acceleration $a$ is given by: ($g$ is acceleration due to gravity)

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$A$ block is projected with speed $20 \,m/s$ on a rough horizontal surface. The coefficient of friction $(\mu)$ between the surfaces varies with time $(t)$ as shown in the figure. The speed of the body at the end of $4 \,s$ will be ............ $m/s$ $(g = 10 \,m/s^2)$.

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$A$ rope of length $L$ and mass $M$ is being pulled on a rough horizontal floor by a constant horizontal force $F = Mg$. The force is acting at one end of the rope in the same direction as the length of the rope. The coefficient of kinetic friction between the rope and the floor is $1/2$. Then,the tension at the midpoint of the rope is

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$A$ body is moving along a horizontal surface with a velocity of $4 \,m/s$. If the coefficient of kinetic friction is $0.2$, the distance travelled by the body before coming to rest is $(g = 10 \,m/s^2)$. (in $\,m$)

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The retarding acceleration of $7.35\, ms^{-2}$ due to frictional force stops a car of mass $400\, kg$ travelling on a road. The coefficient of friction between the tyre of the car and the road is

$A$ block of mass $m$ slides along a floor while a force of magnitude $F$ is applied to it at an angle $\theta$ as shown in the figure. The coefficient of kinetic friction is $\mu_{K}$. Then,the block's acceleration $a$ is given by: ($g$ is acceleration due to gravity)

$A$ block is projected with speed $20 \,m/s$ on a rough horizontal surface. The coefficient of friction $(\mu)$ between the surfaces varies with time $(t)$ as shown in the figure. The speed of the body at the end of $4 \,s$ will be ............ $m/s$ $(g = 10 \,m/s^2)$.

$A$ body is moving along a horizontal surface with a velocity of $4 \,m/s$. If the coefficient of kinetic friction is $0.2$, the distance travelled by the body before coming to rest is $(g = 10 \,m/s^2)$. (in $\,m$)

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