How Does Trailer Weight Impact the Work Done by a Car?

AI Thread Summary
The discussion centers on the relationship between trailer weight and the work done by a car in pulling it. The calculation for work done is clarified as force multiplied by displacement, leading to a conclusion of 160,000 J for a 20 N force over 8 km. Participants express confusion about how trailer weight affects work, particularly regarding kinetic energy and speed. It is explained that while the work done remains the same if the force and distance are constant, a heavier trailer experiences a smaller increase in speed due to its greater mass. The conversation also raises the possibility that the force may primarily be used to overcome resistance rather than increase kinetic energy.
influx
Messages
162
Reaction score
1
EDIT: I JUST REALISED YOU CAN'T DO 1 WORD THREAD TITLES, SORRY ! WON'T DO IT AGAIN!

A car pulls a trailer of weight 2500 N with a force of 20 N for a distance of 8 km along
a horizontal road.

How much work is done by the car in pulling the trailer?
A 160 J
B 20 000 J
C 160 000 J
D 20 000 000 J

I know work done = force x displacement

so it should be 20 x 8000 = 160,000 J (so the answer is C)

But what I am confused about is surely the weight of the trailer will affect how much work the car does in pulling it? I mean for instance if the trailer was 5 N, shouldn't the work done be MUCH less .. BTW can you please try explain in simple terms since I am not that advanced at Physics as you guys.

Cheers :)
 
Physics news on Phys.org
hi influx! :smile:
influx said:
A car pulls a trailer of weight 2500 N with a force of 20 N for a distance of 8 km along
a horizontal road.

How much work is done by the car in pulling the trailer?

… surely the weight of the trailer will affect how much work the car does in pulling it? I mean for instance if the trailer was 5 N, shouldn't the work done be MUCH less

work done = change in energy

if you applied the same force for the same distance, the two trailers' increased kinetic energy would be the same

but the heavier trailer's speed would have increased less :wink:
 
tiny-tim said:
..

Hi :)

Oh! Never knew that! When you say change in energy, what exactly do you mean? And does the speed increase less because kinetic energy = 1/2(mv2) , so as the value for 'm' increases, the value of 'v' decreases?

Cheers
 
influx said:
When you say change in energy, what exactly do you mean? And does the speed increase less because kinetic energy = 1/2(mv2) , so as the value for 'm' increases, the value of 'v' decreases?

if the speed goes from u to v, then the change in energy (on a horizontal road) is 1/2mv2 - 1/2mu2

if that is the same for two masses, the one with larger m must have smaller (v2 - u2)
 
Who says that the 20N force is increasing the KE of the trailer, and not just overcoming air resistance?
 
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