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A constant force F is pushing a 5 kg mass on a horizontal surface at a constant velocity of 2m/s. The coefficient of friction between the surface and the mass is 0.3. Take g=10m/s^2. If F acts along the direction of motion, the rate at which F is doing work (in watts) is : a)3, b)6, c)10, d)15, e)30 (answer) How do you get this answer? Thanks!


4 years ago

## Answers : (1)

Raman Mishra
67 Points
							Rate of doing work is known as power and is given by the equation:               $\dpi{80} P = F.v = Fv\cos \theta$  where $\dpi{80} \theta$ is the angle between F and v.Here since F is along the direction of motion therefore $\dpi{80} \cos \theta = \cos0^{0} =1$ and $\dpi{80} P = Fv$Now, since there is no acceleration the applied force F is used in counteracting the frictional force which is given by:              $\dpi{80} f = \mu N$  i.e, $\dpi{80} F=f$Substituting $\dpi{80} \mu = 0.3$ and $\dpi{80} N= mg = 5\times 10 = 50N$ , we get              $\dpi{80} f = 15N$           $\dpi{80} = >$ $\dpi{80} F=15N$       $\dpi{80} \therefore$  $\dpi{80} P = Fv = 15\times 2 = 30watts$

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