A block of mass m is directly pulled up slowly on a smooth inclined plane of height h and inclination θ with the help of a string parallel to the incline. Among the following statements, the one that is incorrect for the block when it moves up from the bottom to the top of the incline is

Block is slowly moved it means net acceleration is zero, so

$\mathrm{T}=\mathrm{mgsin}\theta$

(i) W_N = 0, because normal reaction is perpendicular to displacement.

(ii) ${\mathrm{W}}_{\mathrm{T}}=\mathrm{mgsin}\mathrm{\theta }\times \left(\frac{\mathrm{h}}{\sin \mathrm{\theta }}\right)\times \cos 0^{\circ }= \mathrm{mgh}$

(iii) W_g = mg x component of displacement along mg

So ${\mathrm{W}}_{\mathrm{g}}=\mathrm{mg}(-\mathrm{h})=-\mathrm{mgh}$

(iv) By work energy theorem

${\mathrm{W}}_{\text{All }}=\mathrm{\Delta KE}$

But, $\mathrm{\Delta }KE=0,\text{ So }{\mathrm{W}}_{\mathrm{All}}=0$