Calculating Horsepower and Power Formula for 1000Kg Auto on 3% Grade at 20m/s

  • Thread starter Thread starter SS2006
  • Start date Start date
  • Tags Tags
    Formula Power
AI Thread Summary
To calculate the horsepower required for a 1000Kg auto traveling up a 3% grade at 20m/s, the approach should focus on the power formula, which is power equals force times velocity. Since the car is moving at a constant speed, there is no acceleration involved, and the only force to consider is the gravitational component acting against the slope. The resistive force is determined by the weight of the vehicle and the slope's angle, which can be calculated using trigonometry. It's important to use the vertical component of the speed to find the effective force. Understanding gradients in trigonometry is essential for accurate calculations in this scenario.
SS2006
Messages
80
Reaction score
0
A 1000Kg auto travels up a 3 percent grade at 20m/s. Find the horsepower required, neglection friction.
ok HP, that means power formula
work/time right, or f * v
how do i approach here
i wanted to do v2 squre = v1 square + 2 ad. but i dotn know if v1 is 0 and is distance (3/100)?
i did that and i got a weird acceleration number, plugged that into f = ma, got force and mulitplied that by V which is 20 and i got the wrong answer, any ideas?
thanks
 
Physics news on Phys.org
Wrong direction. There is no acceleration here. The 20 m/s is implied to be constant (it never says "starts at rest" or the like, only "travels at 20 m/s).

power is F*v as you noticed. In this situation, the force used to maintain constant speed is equal to the resistave forces. Since there is no friction, the only resistive force is the component of weight that points downhill, parallel to the slope.

Or (to do the same thing differently) you could take the full weight at the resistive force and multiply this by the vertical componant of the speed (how much of the 20 m/s points straight up vertically?)

Have you done trigonometry using "gradients" instead of degrees?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top