Minimum Horsepower for Dragging a Box at a Given Speed

  • Thread starter Thread starter engineer2010
  • Start date Start date
  • Tags Tags
    Power
AI Thread Summary
To determine the minimum horsepower required to drag a 310 kg box at 1.2 m/s on a level floor with a coefficient of friction of 0.45, the frictional force (Ff) was calculated to be 1367 N. The participant initially struggled with understanding the force exerted by the motor, mistakenly considering additional forces. After clarification, they realized the only force needed was the frictional force, leading to the correct application of the power formula. The conversion from watts to horsepower was also highlighted as a necessary step. Ultimately, the participant successfully solved the problem.
engineer2010
Messages
10
Reaction score
0
Quick help with Power problem!

Homework Statement


What is the minimum horse power a motor must have to be able to drag a 310 kg box along a level floor at a speed of 1.2 m/s if the coefficient of friction is .45


Homework Equations


fg=mg - - - > (c of f) (fn)= Ff
P = Fv



The Attempt at a Solution



I found Ff to be 1367 N but I am unsure of how to find the force with the motor is putting on the box.. I know that the box is not acclerating. so i tried going -1367 +F= 0 and the putting that in the power formula but that did not give me the correct answer
 
Physics news on Phys.org
Are you remembering to convert it from watts to horsepower?
 
ok i got it thanks.. i was thinking that there was another force in addition to the 1367. but i figured it out.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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