Question

In the figure shown, match the following two columns. ($g=10m{s}^{-2}$.)

Answer 1

A.

As the system is accelerating at the rate of $a=5m{s}^{-2}$, the system experiences pseudo force $P=2kg\times g=10N$ (rightward).

Thus the total normal force on the surface is $N=10N+10N=20N\left(leftward\right)$.

B.

Weight of the block $W=mg=2kg\times g=20N$ (downward).

When $F=15N$ (upward) acts on the block, the opposing force $O=15N\left(upward\right)+20N\left(downward\right)=5N\left(downward\right)$

Frictional force $f={\mu }_k\times N$ (upward, since the block is in downward motion)

$\Rightarrow f_{max}=(0.3\times 20)N=6N$ (upward)

Note: Frictional force value never exceeds the opposing force value. It always tries to get even with the opposing force (if f’s upper limit > opposing force).

Here $f_{max}>O\Rightarrow f=5N$

C.

We need to find the minimum value of $F$ to stop the block from moving down.

$\Rightarrow$ relative motion is downward, hence $f$ must act upward.

Therefore, $F+f=mg\Rightarrow maximisingfminimisesF$

$F+6N=20N\Rightarrow F=14N\left(upward\right)$

D.

Minimum force required to stop the block from moving up is $F=0N$ as the block never moves upward in its natural state unless external force is applied.

Thus, $A⟶4B⟶1C⟶4D⟶4$