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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!

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!

Grade:11

1 Answers

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

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