$A$ bullet of mass $20 \,g$ moving with $500 \,ms^{-1}$ is pierced $1 \,cm$ into a wooden block. The retarding force experienced by the bullet is:

The velocity of an object of mass $2 \ kg$ is given by $v = (8t \hat{i} + 3t^2 \hat{j}) \ m/s$, where $t$ is time in seconds. What will be the direction of the net force on the object relative to the positive direction of the $X$-axis, at the instant when its magnitude is $20 \ N$?

$A$ force $\vec{F}=(40 \hat{i}+10 \hat{j}) \, N$ acts on a body of mass $5 \, kg$. If the body starts from rest,its position vector $\vec{r}$ at time $t=10 \, s$ will be -

$A$ body of $5 \, kg$ is moving with a velocity of $20 \, m/s$. If a force of $100 \, N$ is applied on it for $10 \, s$ in the same direction as its velocity,what will be the final velocity of the body in $m/s$?

At any instant,the velocity of a particle of mass $500 \, g$ is $(2t \hat{i} + 3t^2 \hat{j}) \, ms^{-1}$. If the force acting on the particle at $t = 1 \, s$ is $(\hat{i} + x \hat{j}) \, N$,then the value of $x$ will be:

$A$ body of mass $2 \ kg$ moving with velocity of $\overrightarrow{v}_{in} = 3 \hat{i} + 4 \hat{j} \ ms^{-1}$ enters into a constant force field of $6 \ N$ directed along the positive $z$-axis. If the body remains in the field for a period of $\frac{5}{3} \ s$, then the velocity of the body when it emerges from the force field is: