Force, speed of truck on a 10.0 degree hill

In summary, the truck can maintain a constant speed on a hill with a slope of 10.0° if it delivers a total of 3347.8 newtons of power.
  • #1
Valerie Prowse
25
0

Homework Statement



A loaded truck has a mass of 3100 kg. The maximum speed it can maintain on a 5.0° hill is 80 km/h. What constant speed could the truck maintain on a hill with a slope of 10.0°? Assume the total force due to air resistance and friction is 700 N and that it does not vary with speed.

Homework Equations



∑F = ma

The Attempt at a Solution



I have an exam tomorrow and this is one of the practice questions. The answer is 44.8 km/h, but I have no idea how to get there. What I have worked out so far is the FBD of the truck, and:
since a = 0, for θ = 10
∑F = 0
F - Ffr - mg⋅sin10 = 0
F = Ffr + mg⋅sin10
F = 5975.4 N
and vθ=10 = ??

I am having trouble relating this back to a velocity. Also, since there is no variation in air resistance and friction, it would mean that:
for θ = 5
F = Ffr + mg⋅sin5
F = 3347.8 N
and vθ=5 = 22.2 m/s

I feel like I am almost there but I can't quite put the pieces together ... I've though about work and momentum, but those don't seem to fit with the idea of the question ...
Does anyone have an idea on where to go from here??
Thanks!
 
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  • #2
Have you considered thinking in terms of the power the truck can deliver?
 
  • #3
gneill said:
Have you considered thinking in terms of the power the truck can deliver?

I'm not sure what you mean by that..
 
  • #4
Valerie Prowse said:
I'm not sure what you mean by that..
While the truck is climbing the 5° slope it is said to be doing the maximum speed that it can achieve on that slope. So it (or rather its engine) must be delivering the most power it can under the circumstances. Can you determine that power?
 
  • #5
gneill said:
While the truck is climbing the 5° slope it is said to be doing the maximum speed that it can achieve on that slope. So it (or rather its engine) must be delivering the most power it can under the circumstances. Can you determine that power?

Ah, I see the problem. This question is from a previous exam, but my course did not cover this topic, hence why I had no idea what you meant by power. I went through my textbook and found the section on power, which was not assigned, but I will give it a quick look over and try this question again later.
Thank you for your help, though!
 

1. What is the formula for calculating the force on a truck going up a 10.0 degree hill?

The formula for calculating force is F = mgsinθ, where m is the mass of the truck, g is the acceleration due to gravity (9.8 m/s²), and θ is the angle of the hill (10.0 degrees in this case).

2. How does the force on the truck change as it goes up the hill?

The force on the truck will increase as it goes up the hill because the angle θ increases, resulting in a larger sinθ value in the formula. This means that the truck will require more force to overcome the gravitational force pulling it down the hill.

3. What is the relationship between the speed of the truck and the force needed to climb the 10.0 degree hill?

The greater the speed of the truck, the greater the force needed to climb the hill. This is because the truck will have more kinetic energy, and therefore, more force is required to overcome the force of gravity and maintain its speed.

4. How does the mass of the truck affect the force needed to climb the hill?

The mass of the truck has a direct relationship with the force needed to climb the hill. The greater the mass, the greater the force needed to overcome gravity and maintain the same speed.

5. Is there a limit to the speed a truck can reach while climbing a 10.0 degree hill?

There is no specific limit to the speed a truck can reach while climbing a hill, as it depends on factors such as the truck's engine power, weight distribution, and road conditions. However, the force needed to maintain a certain speed will increase as the incline of the hill increases, so there may be a point where the truck's engine power is not enough to overcome the force of gravity and maintain its speed.

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