What is the average pulling force?

[kurtlau]

COULD ANYONE PLEASE ANSWER ME FOR I DON'T UNDERSTAND WHY SHOULDN'T WE CONSIDER mgsin(angle) and friction AS WELL?

At the top of a slope, a trolley(mass:20kg) is released from rest and slides down freely along the slope (the runway is 25m long and makes an angle of 15 degrees with the horizontal, the frictional force is 20N).
1. Find the acceleration of the trolley.(1.59)
2. Find the speed of the trolley when it is half way down the slope.(6.3)

Then, a person catches the trolley when it is half way down the slope and makes it stop at the foot of the slope. What is the AVERAGE PULLING FORCE?
v=0,u=6.3,s=12.5
Since v^2-u^2=2as,
a= -1.59
Average pulling force=20(1.59)=31.8
I DON''T UNDERSTAND WHY WE NEEDN'T HAVE
Net force = ma --> Pulling force-mgsin15+f=20(-1.59)

THANK YOU VERY MUCH!

 

[jdavel]

kutlau,

It looks to me as though you're confused on a few points.

I think you need to consder friction in your calculation for the speed the trolley is going when it's halfway down the ramp (the part where no one is pulling) Edit: On second look, it seems you're doing this correctly.

I'm pretty sure your'e supposed to assume that the person pulls parallel to the ramp. So do you need that sin(15) factor?

 

[Doc Al]

Makes sense to me!

COULD ANYONE PLEASE ANSWER ME FOR I DON'T UNDERSTAND WHY SHOULDN'T WE CONSIDER mgsin(angle) and friction AS WELL?

Of course you should.

At the top of a slope, a trolley(mass:20kg) is released from rest and slides down freely along the slope (the runway is 25m long and makes an angle of 15 degrees with the horizontal, the frictional force is 20N).
1. Find the acceleration of the trolley.(1.59)

Right. (Using a value of g = 10 m/s^2.)

2. Find the speed of the trolley when it is half way down the slope.(6.3)

Right.

Then, a person catches the trolley when it is half way down the slope and makes it stop at the foot of the slope. What is the AVERAGE PULLING FORCE?
v=0,u=6.3,s=12.5
Since v^2-u^2=2as,
a= -1.59

Right. Acceleration is equal in magnitude, but now points up the slope. (The sign convention you are using is down = positive.)

Average pulling force=20(1.59)=31.8

This is the average net force on the trolley, not the pulling force of the person.

I DON''T UNDERSTAND WHY WE NEEDN'T HAVE
Net force = ma --> Pulling force-mgsin15+f=20(-1.59)

Assuming that by "pulling force" they mean the average force exerted by the person (parallel to the slope), then you are on the right track but made an error in sign. The pulling force (F) and the friction both point up the slope:
-F +mgsin(15) -20 = ma = 20(-1.59), so
F = mgsin(15) -20 +20(1.59) = 63.6 N up the slope
This makes sense. As it rolls down, there is a net force of 31.8N down the slope, so to make it stop with the same acceleration (but negative) you have to have a net force of 31.8N up the slope. Which means you have to push up with a force of 63.6N.

 

[kurtlau]

thank you very much