Work Done on a Suitcase by Various Forces

  • Thread starter Thread starter Crusaderking1
  • Start date Start date
  • Tags Tags
    Work
AI Thread Summary
A luggage handler exerts a force of 155 N on a 16.7 kg suitcase up a 30-degree ramp, resulting in various work calculations. The work done by the applied force is calculated as 550 J, while the work done by friction is -179 J. The normal force and gravitational force do not contribute to work since their angles relative to the displacement are 90 degrees, resulting in zero work. The total work done on the suitcase is 371 J. The discussion concludes with a request for assistance in determining the suitcase's speed after traveling 4.10 m along the ramp.
Crusaderking1
Messages
158
Reaction score
0

Homework Statement



*I already figured the problem out*

A luggage handler pulls a suitcase of mass 16.7 kg up a ramp inclined at an angle 30.0 degrees above the horizontal by a force \overline{F} of magnitude 155 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.267. The suitcase travels a distance 4.10 m along the ramp.

A. Calculate the work done on the suitcase by the force F.
B. Calculate the work done on the suitcase by the gravitational force.
C. Calculate the work done on the suitcase by the normal force.
D. Calculate the work done on the suitcase by the friction force.
E. Calculate the total work done on the suitcase.

F. If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 4.10 m along the ramp?

Homework Equations



F*s = Work

The Attempt at a Solution



A. Is it:
work of F applied= (155 N)(4.10 m)cos30.0 = 550.0 J

B. Is it
work of friction=(43.7 N)(4.10 m)cos180.0 = -179.0 J

C and D. Wouldn't both be 0, considering cos90 = 0?

E. Is it just 550 J + 0 J + 0 J + -179 J = 371 J?

F. No idea, need help.

Thoughts? Thanks!
 
Last edited:
Physics news on Phys.org
s is along the ramp right? So, why are you projecting things to the horizontal direction?
 
So work of f applied= 155 N(4.10 m)?

*I figured this problem out*
 
Last edited:
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