148733 A Block of mass $50 \mathrm{~kg}$ is pulled at a constant speed of $4 \mathrm{~m} \mathrm{~s}^{-1}$ across a horizontal floor by an applied force of $500 \mathrm{~N}$ directed $30^{\circ}$ above the horizontal. The rate at which the force does work on the block in watt is
148734
A particle of mass $500 \mathrm{gm}$ is moving in a straight line with velocity $\mathbf{v}=b x^{5 / 2}$. The work done by the net force during its displacement from $x=0$ to $x=4 \mathrm{~m}$ is :
$\text { (Take } b=0.25 \mathrm{~m}^{-3 / 2} \mathrm{~s}^{-1} \text { ) }$
148733 A Block of mass $50 \mathrm{~kg}$ is pulled at a constant speed of $4 \mathrm{~m} \mathrm{~s}^{-1}$ across a horizontal floor by an applied force of $500 \mathrm{~N}$ directed $30^{\circ}$ above the horizontal. The rate at which the force does work on the block in watt is
148734
A particle of mass $500 \mathrm{gm}$ is moving in a straight line with velocity $\mathbf{v}=b x^{5 / 2}$. The work done by the net force during its displacement from $x=0$ to $x=4 \mathrm{~m}$ is :
$\text { (Take } b=0.25 \mathrm{~m}^{-3 / 2} \mathrm{~s}^{-1} \text { ) }$
148733 A Block of mass $50 \mathrm{~kg}$ is pulled at a constant speed of $4 \mathrm{~m} \mathrm{~s}^{-1}$ across a horizontal floor by an applied force of $500 \mathrm{~N}$ directed $30^{\circ}$ above the horizontal. The rate at which the force does work on the block in watt is
148734
A particle of mass $500 \mathrm{gm}$ is moving in a straight line with velocity $\mathbf{v}=b x^{5 / 2}$. The work done by the net force during its displacement from $x=0$ to $x=4 \mathrm{~m}$ is :
$\text { (Take } b=0.25 \mathrm{~m}^{-3 / 2} \mathrm{~s}^{-1} \text { ) }$
148733 A Block of mass $50 \mathrm{~kg}$ is pulled at a constant speed of $4 \mathrm{~m} \mathrm{~s}^{-1}$ across a horizontal floor by an applied force of $500 \mathrm{~N}$ directed $30^{\circ}$ above the horizontal. The rate at which the force does work on the block in watt is
148734
A particle of mass $500 \mathrm{gm}$ is moving in a straight line with velocity $\mathbf{v}=b x^{5 / 2}$. The work done by the net force during its displacement from $x=0$ to $x=4 \mathrm{~m}$ is :
$\text { (Take } b=0.25 \mathrm{~m}^{-3 / 2} \mathrm{~s}^{-1} \text { ) }$