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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!

Grace Mathew , 8 Years ago

Grade 11

1 Answers

Raman Mishra

Last Activity: 8 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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