Question

A block of mass $2.0\text{kg}$ is pulled up on a smooth incline of angle ${30}^{\circ }$ with the horizontal. If the block moves with an acceleration of $1.0{\text{m/s}}^2$, then the power delivered by the pulling force at a time $4.0\text{s}$ and the average power delivered during the $4.0\text{s}$ after the motion starts are

• A. $36W,24W$
• B. $24W,24W$
• C. $24W,12W$
• D. $48W,24W$

Answer 1

The correct option is D $48W,24W$


As we konw that,
$P=\overrightarrow{F}.\overrightarrow{v}$
$P=Fv\cos \theta$
$\theta$ is the angle between force and velocity
In this case $\theta =0$ then $\cos \theta =1$
So, $P=Fv$
as, $v=u+at=0+1\times 4$
$v=4m/s$

For finding $F$ resolving force in verticle direction
$F-mg\sin \theta =2$
$F-2\times 10\times \frac{1}{2}=2$
$F=12N$
So, Power $\left(P\right)=Fv=48W$

Part 2:
$P_{avg}=\frac{W_{final}-W_{initial}}{\Delta t}$
but, $W_{initial}=0$
As $W_{final}=F.s$(displacement)
From second law, $s=ut+\frac{1}{2}a{t}^2$
$s=0+\frac{1}{2}\times 1\times 4^2=8m$
$p_{avg}=\frac{12\times 8}{4}=24W$