How Much Work is Done Pushing a Crate Up a Plank?

  • Thread starter Thread starter lilmissbossy
  • Start date Start date
  • Tags Tags
    Mass Work
AI Thread Summary
The discussion focuses on calculating the work done in pushing a crate up a plank and the increase in potential energy (PE) once the crate is on a platform. The work done is calculated using the formula W=Fxd, resulting in 2450 N. The increase in potential energy is determined using PE=mgh, yielding a value of 1176 N. The calculations assume no friction and that the force is applied along the length of the plank. The participant expresses satisfaction with their correct solution.
lilmissbossy
Messages
14
Reaction score
0

Homework Statement


force F is required to push a crate of mass m up a plank of length L into a truck whose platform is a vertical distance h above the road.

a)How much work is done pusing the crate up the plank m=100kg, L=5m, F=490N h=1.2m
b)what is the increase in PE of the crate once the crate is on the platform


Homework Equations


W=Fxd
PE=mgh



The Attempt at a Solution



ok so my working is

W=Fxd
W=490Nx5m
W=2450N

and

PE=mgh
PE=(100kg)(9.8ms-1)(1.2m)
PE=1176 N

Hmmmmmmm...i think they are right
 
Physics news on Phys.org
Yeah it looks ok, provided no friction is involved and F is directed along the length of L.
 
Excellent i finally got one right by myself!
 
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}...
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...
Back
Top