A block of mass m is moving on a rough horizo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The surface exerts two forces on the block: the normal force $(N)$ acting vertically upwards and the kinetic frictional force $(f_{k})$ acting hor

Vedclass · Accepted Answer

$mg(1+\mu_{k}^{2})^{1/2}$

Vedclass · Answer

(A) The surface exerts two forces on the block: the normal force $(N)$ acting vertically upwards and the kinetic frictional force $(f_{k})$ acting horizontally opposite to the direction of motion.$1$. The normal force is $N = mg$.$2$. The kinetic frictional force is $f_{k} = \mu_{k} N = \mu_{k} mg$.$3$. The net force $(F_{net})$ exerted by the surface is the vector sum of the normal force and the frictional force:$F_{net} = \sqrt{N^{2} + f_{k}^{2}}$$F_{net} = \sqrt{(mg)^{2} + (\mu_{k} mg)^{2}}$$F_{net} = \sqrt{(mg)^{2} (1 + \mu_{k}^{2})}$$F_{net} = mg \sqrt{1 + \mu_{k}^{2}}$$F_{net} = mg(1 + \mu_{k}^{2})^{1/2}$Therefore,option $A$ is correct.

$A$ block of mass $m$ is moving on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is $\mu_{k}$. The net force exerted by the surface on the block is (where $g$ is the acceleration due to gravity).

Similar Questions

$A$ man slides down a telegraphic pole with an acceleration equal to one-fourth of the acceleration due to gravity. The frictional force between the man and the pole is equal to which of the following in terms of the man's weight $w$?

$A$ small box resting on one edge of the table is struck in such a way that it slides up to the other edge,$1 \, m$ away,after $2 \, s$. The coefficient of kinetic friction between the box and the table

Which one of the following statements is correct?

$A$ body of mass $m$ moves with a velocity $v$ on a surface whose friction coefficient is $\mu$. If the body covers a distance $s$ before coming to rest,what is the initial velocity $v$?

$A$ block of mass $2 \,kg$ is being pushed against a wall by a force $F=90 \,N$ as shown in the figure. If the coefficient of friction is $0.25$, then the magnitude of acceleration of the block is (Take $g=10 \,ms^{-2}$, $\sin 37^{\circ}=\frac{3}{5}$). (in $\,ms^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module