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under the action of a force, 2 kg body moves such that its position x as a function of time is given by x = (t)^(2/3) where x is in metres and t is in seconds. the work done by the force in the first two seconds is ?
under the action of a force, 2 kg body moves such that its position x as a function of time is given by x = (t)^(2/3) where x is in metres and t is in seconds. the work done by the force in the first two seconds is ?

```
4 years ago

Vikas TU
14149 Points
```							velocity = dx/dt = (2/3)*t^-1/3acceleration = (-2/9)*t^-4/3Put t=2 in both v and accln.Work =  Force x VelocityF=ma and velocity has been calculated already.
```
4 years 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