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`        This question is from HC Verma.I just have a doubt here.Here work done is calculated by formula F×d.Why we can't use W=Change in kinetic energy here.I used this to solve work done bit can't get the answer.So,can you please solvw the whole question using W=change in kinetic energy or tell me where I was wrong.`
10 months ago

Kapil Khare
80 Points
```							In this question the work done on cart to pull it up is not only changing the kinetic energy of the cart but some some work done is also increasing the potential energy of the cart. So, we also have to consider the change in the potential energy.velocity of cart after 4 seconds = v = at = (1m/s2)(4s) = 4m/sdisplacement of cart = s = at2/2 = (1m/s2)(4s)2/2 = 8mheight at which cart reached after 4 seconds = h = s(sin30$\degree$) = 8*(0.5) = 4mSo, work done by the force = W = (mv2/2) + (mgh) – 0 = [(2kg)(4m/s)2(0.5) + (2kg)(9.8m/s2)(4m)$\implies$                                       W = (16 + 78.4)Joules$\implies$                                       W = 94.4 J Power = P = W/t = (94.4J)/(4s)$\implies$      P = 23.6Watt
```
10 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions