Question

Length of an inclined plane is $5$ m and it is inclined at ${30}^{\circ }$ to the horizontal. Work done to move the block up the plane with uniform velocity is $100$ J. Then work done to move the block up the plane with uniform acceleration of $4m{s}^{-2}$ is $(\tan {30}^{\circ }=1/\sqrt{3})$

• A. $180J$
• B. $400J$
• C. $300J$
• D. $100J$

Answer 1

The correct option is A $180J$

For the body to move with uniform speed, resultant force on the body should be zero.
Therefore, the force applied on the body is equal to the weight force of the body along the plane.
$F=mg\sin \theta$
$100=m\left(10\right)\sin {30}^{\circ }\left(5\right)$
$m=\frac{100}{\left(10\right)\left(0.5\right)\left(5\right)}=4kg$
Now the force required to move the block up the plane with an acceleration of 4$m{s}^{-2}$ is
$F-mg\sin \theta =ma$
$F=m(a+g\sin \theta )=4(4+10(0.5\left)\right)=36N$
Work done$=36\times 5=180J$