Dear Student, i) here,Force apllied,F=mg=40×10=400Ndistance covered in the direction of force,d=63×15 cm=9 45 mwe know,work,W=FdW=400×9 45=3780Jpower=worktime=378018=210J/s=210 watt kindly ask different question in different post Regards

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Dear Student,
i)

h e r e, F o r c e a p l l i e d, F = m g = 40 \times 10 = 400 N d i s \tan c e c o v e r e d i n t h e d i r e c t i o n o f f o r c e, d = 63 \times 15 c m = 9.45 m w e k n o w, w o r k, W = F d W = 400 \times 9.45 = 3780 J p o w e r = \frac{w o r k}{t i m e} = \frac{3780}{18} = 210 J / s = 210 w a t t

kindly ask different question in different post.

Regards

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