The frictional force acting on 1 ,kg block is | vedclass

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

Question

(a)

If both move together $a=\frac{10}{101} \simeq 0.1 \,m / s ^2$

Now, $F_{	ext {net }}=1(0.1)=0.1 \,N$

$f_L=(0.5)(1)(g)=5 \,N$

So, $f=0

Vedclass · Accepted Answer

$0.1$

Vedclass · Answer

(a)

If both move together $a=\frac{10}{101} \simeq 0.1 \,m / s ^2$

Now, $F_{	ext {net }}=1(0.1)=0.1 \,N$

$f_L=(0.5)(1)(g)=5 \,N$

So, $f=0.1 \,N$

The frictional force acting on $1 \,kg$ block is .................. $N$

Similar Questions

A rod $(AB)$ is attached to a fixed point $(C)$ using a light rope $(AC)$. The other end of the rod $(B)$ is sitting on ice with negligible friction and the system is in stationary position. Which of the following can be the equilibrium configuration of this system?

A block is moving on an inclined plane making an angle $45^{\circ}$ with the horizontal and the coefficient of friction is $\mu$. The force required to just push it up the inclined plane is $3$ times the force required to just prevent it from sliding down. If we define $N=10 \ \mu$, then $N$ is

A block pressed against the vertical wall is in equilibrium. The minimum coefficient of friction is:-

With what minimum velocity should block be projected from left end $A$ towards end $B$ such that it reaches the other end $B$ of conveyer belt moving with constant velocity $v$. Friction coefficient between block and belt is $\mu$ .

A block weighs $W$ is held against a vertical wall by applying a horizontal force $F$. The minimum value of $F$ needed to hold the block is $[\mu < 1]$