1. ## Physics Help!!

hi anyone can help me on this?

(a) A 3 ton vehicle travels along a horizontal road at 50km/h against a tractive resistance of 150N per ton. Calculate the braking force needed to stop the vehicle in 50m

(b) This vehicle now rolls downhill from rest along a slope of 1 in 20 from a distance of 200m. What is then the velocity assuming the same tractive resistance.

(c) If a braking force of 3kn is then applied. Calculate the distance travelled before coming to rest.

i tried to do this question and not sure bout the answers.
Ans (a) = 1022N
Ans (b) = 11.622
Ans (c) = don't know what to do??

pls can someone help? its really urgent!! =(

2. Originally Posted by danielwu
hi anyone can help me on this?

(a) A 3 ton vehicle travels along a horizontal road at 50km/h against a tractive resistance of 150N per ton. Calculate the braking force needed to stop the vehicle in 50m
First convert every thing to a consistent set of units.

Are the tons long tons or metric tons? I will assume metric tons of 1000 kg.
So the mass of the vehicle is 3000 kg.

50 km/hr = 13.89 m/s

Assume that the total force during the decelleration is constant.

Then the KE of the vehicle is 0.5 3000 13.89^2,
If we have a constant decellerating force f it does work 50 f, which
is equal to the initial KE, so:

50 f = 1500 13.89^2,

or: f=30 13.89^2= 5788.0 N

So the breaking force is this less the tractive resistance of 150x3 N or
5338 N.

RonL

3. Originally Posted by danielwu
hi anyone can help me on this?

(a) A 3 ton vehicle travels along a horizontal road at 50km/h against a tractive resistance of 150N per ton. Calculate the braking force needed to stop the vehicle in 50m

(b) This vehicle now rolls downhill from rest along a slope of 1 in 20 from a distance of 200m. What is then the velocity assuming the same tractive resistance.
Part (b) can be done the same way as part (a) that is work done by force
equals change in energy.

Resolve the gravitational force into two components, one down the plane
and the other normal to the plane. I will also make small angle approximations
so that the angle of the plane is 0.05 radian.

f_d=0.05 mg - 450

so the work done is 200(0.05 mg -450)~=29340 watts...(1)

The change in energy due to work done is gain in KE minus loss of PE = 0.5 m v^2 - 10 mg~=1500 v^2-294300 watts ....(2)

Now equate (1) and (2) and solve for v.

RonL

4. Originally Posted by danielwu

(c) If a braking force of 3kn is then applied. Calculate the distance travelled before coming to rest.
You know the speed of the vehicle when it reaches the end of the slope form part (b).
Hence you know its KE, then if the stopping distance is d the work done is (3000+450)d,
which you equate to the KE and solve for d.

RonL

5. Originally Posted by danielwu
hi anyone can help me on this?

(a) A 3 ton vehicle travels along a horizontal road at 50km/h against a tractive resistance of 150N per ton. Calculate the braking force needed to stop the vehicle in 50m

(b) This vehicle now rolls downhill from rest along a slope of 1 in 20 from a distance of 200m. What is then the velocity assuming the same tractive resistance.

(c) If a braking force of 3kn is then applied. Calculate the distance travelled before coming to rest.

i tried to do this question and not sure bout the answers.
Ans (a) = 1022N
Ans (b) = 11.622
Ans (c) = don't know what to do??

pls can someone help? its really urgent!! =(

This was urgent then. So these answers may be late already. Nevertheless, I continue. For practice.

My way is through the old reliable F = ma. Force = mass*acceleration.
If there is an acceleration (or deceleration), then there mhust be a force causing it.

--------------------------
(a) A 3 ton vehicle travels along a horizontal road at 50km/h against a tractive resistance of 150N per ton. Calculate the braking force needed to stop the vehicle in 50m

F = ma.
So we need to determine forces, mass, and acceleration or deceleration.
If the vehicle is mving, then it is brought to a stop, then there must be a deceleration.
(final velocity minus initial velocity) / (time) = acceleration.
(Vf -Vo)/t = a
Vo = 50km/h*(1000m/1km)*(1hr/3600sec) = 50(1000)/3600 = 13.889 m/sec.
So,(0 -13.889)/t = a --------------------(i)

We don't know t, the time in seconds that will take the vehicle to travel 50 meters to stop after applying the brakes.
(Average velocity)*(time) = distance.
(13.889 +0)/2 *t = 50
t = 50*2/13.89 = 7.2 sec --------***

Substitute that into (i),
(-13.889)/(7.2) = a
a = -1.929 m/sec/sec ------------***

The force causing this deceleration is the braking force B, plust the tractive resistance 3(150N) = 450N.
So,
F = ma
B +450 = (3*1000)(1.929)
B = 5787 -450

----------------------------------------------------
(b) This vehicle now rolls downhill from rest along a slope of 1 in 20 from a distance of 200m. What is then the velocity assuming the same tractive resistance.

Draw the figure on paper.
The force due to weight is vertically downwards, W = (3*1000)(9.8) = 29,400 newtons.
The component of W that is parallel to the incline is 1468.166 newtons.
[I assume you know how to get that. It is by proportion, x/1 = 29,400/sqrt(401). x=1468.166]

So the forces involved are 1468.166N going down and 3*150 = 450N going up. The one going down is greater than the one going up, so the vehicle will roll and accelerate downwards.
F = ma
(1468.166 -450) = (3*1000)(a)
1018.166 = 3000a
a = 1018.166 / 3000 = 0.339 m/sec/sec ------------***

So the vehicle is accelerating at 0.339 m/sec/sec downwards. When it has travelled 200m from stop, what is its velocity that instant?
(Vf -Vo)/t = a
(Vf -0)/t = 0.339
Vf = 0.339*t -----------------(ii)

we don't know t, the time in seconds that took the vehicle to travel that 200m distance.
(average velocity)/2 *t = distance
(0 +Vf)/2 *t = 200
(Vf)t = 400
t = 400/Vf ----------------***

Substitute that into (ii),
Vf = 0.339 *400/Vf
(Vf)^2 = 0.339*400 = 135.6
Vf = sqrt(135.6) = 11.645 m/sec --------------answer.

-----------------------------------------------------------
(c) If a braking force of 3kn is then applied. Calculate the distance travelled before coming to rest.

3kn = 3k N = 3000 newtons.

F = ma
(1018.166 -3000 -450) = (3*1000)a
-2431.834 = 3000a
a = -2431.834 /3000 = -0.811 m/sec/sec --------***

(Vf -Vo)/t = a
(0 -11.645)/t = -0.811
t = (-11.645) /(-0.811)
t = 14.359 sec -------------------------------***

(Vo +Vf)/2 *t = distance
(11.645 +0)/2 *14.359 = distance
distance = (11.645)/2 *14.359 = 83.6 m

Therefore, the vehicle travelled 83.6 meters before it stopped. ------------answer.