Recent content by mrsteve

I thought you needed it because theta is the angle between the force vector and the velocity vector. No? It changes the answer a lot without it. 90*50 = 4.5*10^3 I got 4.5*10^2

What about F_biker=mg\sin\theta => (75)(9.8)(\sin7)=9.0*10^1 P=(90)(5.0)(\cos7)=4.47*10^2 => 4.5*10^2 = 450W

Potential energy isn't covered until the next chapter. Also, I don't know the height, so how would I find PE?

Homework Statement A bicyclist coasts down a 7.0 degree slope at a steady speed of 5.0 m/s. Assuming a total mass of 75 kg (bicycle plus rider), what must the cyclist's power output be to pedal up the same slope at the same speed? Homework Equations P=Fvcos\theta K=\frac{1}{2}mv^2...

The book has 2,000 J. I suppose that's because there's only 1 significant digit in mass = 1kg? 1.6 *10^3 => 2 *10^3

Also, if the speed of the mass is 2 m/s @ x=2 what is the speed at x=6? KE=\frac{mv^2}{2} 1600=\frac{(1)(v^2)}{2} => v=5.6*10^1 = 60 m/s Would that be correct?

Homework Statement A force given by F(x)=5x^3 (in \frac {N}{m^3}) acts on a 1-kg mass moving on a frictionless surface. The mass moves from x = 2m to x = 6m. a) How much work is done by the force? Homework Equations W=\int F_x from x_o to x_f The Attempt at a Solution \int 5x^3 from...

Ahh... So 90 km/h = 25 m/s. \frac{1}{2}(7000)(25)^2=2.1875*10^6 => 2.1875*10^6(2) = 4.375*10^6 => 4*10^6 Thanks!

Homework Statement Two railroad cars, each with a mass of 7,000kg and traveling at 90 km/h collide head on and come to rest. How much mechanical energy is lost in this collision? Homework Equations KE=\frac{1}{2}mv^2 The Attempt at a Solution...