A body of weight 64 N is pushed with just en | vedclass

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

Question

(A) Given: Weight of the body,$W = 64 \ N$. Coefficient of static friction,$\mu_s = 0.8$. Coefficient of dynamic (kinetic) friction,$\mu_d = 0.6$. The

Vedclass · Accepted Answer

$0.2 \ g$

Vedclass · Answer

(A) Given: Weight of the body,$W = 64 \ N$. Coefficient of static friction,$\mu_s = 0.8$. Coefficient of dynamic (kinetic) friction,$\mu_d = 0.6$. The force $F$ required to just start the motion is equal to the limiting friction: $F = \mu_s 	imes W = 0.8 	imes 64 \ N$. Once the body starts moving,the kinetic friction acting on it is $f_k = \mu_d 	imes W = 0.6 	imes 64 \ N$. The net force acting on the body is $F_{net} = F - f_k = (0.8 	imes 64) - (0.6 	imes 64) = 64(0.8 - 0.6) = 64 	imes 0.2 \ N$. Using Newton's second law,$F_{net} = m 	imes a$,where mass $m = \frac{W}{g} = \frac{64}{g}$. Substituting the values: $\frac{64}{g} 	imes a = 64 	imes 0.2$. Therefore,$a = 0.2 \ g$.

$A$ body of weight $64 \ N$ is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are $0.8$ and $0.6$ respectively,then the acceleration of the body will be

Similar Questions

$A$ heavy box is pushed across a rough floor with an initial speed of $4 \, m/s$. It stops moving after $8 \, s$. If the average resisting force of friction is $10 \, N$,the mass of the box (in $kg$) is:

The coefficient of friction between the block of mass ${M_1}$ and the surface is $\mu$. When the system is released,it moves with acceleration. What mass $m$ should be placed on the block ${M_1}$ so that the system moves with a constant velocity?

$A$ block of mass $10 \ kg$ is moving on a rough surface with a velocity $v = 4 \ m/s$ to the left. An external force $F = 20 \ N$ is applied to the right as shown in the figure. The coefficient of kinetic friction is $\mu = 0.6$. The frictional force acting on the block is ...... $N$.

$A$ fireman of mass $60 \ kg$ slides down a pole. He is pressing the pole with a force of $600 \ N$. The coefficient of friction between the hands and the pole is $0.5$. With what acceleration (in $m/s^2$) will the fireman slide down? (Take $g = 10 \ m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module