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I have a truck which weighs 190 kg and will travel 100m down a 38° slope.

How I can find the trucks speed at the end of the 100m?

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  • $\begingroup$ Related meta question. $\endgroup$ – HDE 226868 Feb 3 '15 at 16:07
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    $\begingroup$ 38 degrees!(?) that is one heck of a slope! $\endgroup$ – George Herold Feb 3 '15 at 17:35
  • $\begingroup$ @George: Not only that, but it's also irrelevant to the problem (assuming a frictionless mechanical system, which I think is implied). $\endgroup$ – Olin Lathrop May 8 '17 at 20:01
  • $\begingroup$ @OlinLathrop It's not irrelevant, perhaps, as assuming "100m down" the slope means 100m of linear travel, then the incline defines the change in height over that distance? If "100m down" means change in altitude, then Fred's answer is fine... (Sorry for the thread resurrection - got here via the "homework questions" Meta) $\endgroup$ – Jonathan R Swift Dec 18 '17 at 17:08
  • $\begingroup$ @Jon: Yes, if "100 m down" actually means along the slope and not down, then you're right. $\endgroup$ – Olin Lathrop Dec 18 '17 at 17:25
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This is a constant acceleration problem on an inclined plane which ignores friction.

The force acting on the truck is the force of gravity. The acceleration the truck will experience will be solely due to gravity, $9.8\text{ m/s}^2$.

Using the 38 degrees for the slope you have to find the component of gravity/acceleration going down the slope.

Given the truck will travel 100 m down the slope and the acceleration in the plane of the slope that you calculated, you have to rearrange the following equation to get the time required to travel the distance:

$$s=ut+0.5at^2$$

The initial velocity, $u$, will be zero if the truck starts rolling from a standing start.

To find the final velocity of the truck you use the following equation

$$v^2=u^2+2as$$

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Fred's answer solves this problem using classical mechanics/kinematics. However, you can also solve this type of problem using an energy approach (which I have always found easier to do - Newton's laws just feel more intuitive).

Newton's first law (conservation of energy)

The baseline potential energy (zero) is the truck's starting position, and the final energy state is the ending position. Assuming the truck comes to a stop at the top, the total energy - all of which is potential energy - of the truck at the ending position will be:

$PE_f=M_{truck}\cdot g\cdot H$

where

$H=D\cdot\tan\theta$

Newton's first law (conservation of energy) says $KE_i = PE_f$ (assuming zero friction), so the formula to find the initial velocity needed for the truck to arrive at the top is:

$KE_i=\frac{1}{2}m v_i^2=PE_f$

Therefore:

$v_{i}=\sqrt{\frac{2PE_f}{m}}$

Newton's second law ($F=ma$)

The time of travel is found using Newton's second law ($F=ma$). The acceleration is the change in velocity over time. The final velocity is zero, so we know the change in velocity from the equation above ($\Delta v=v_i$). We also know the force $F$ acting on the truck is its mass times gravity. Therefore:

$F=ma\rightarrow mg=m\frac{\Delta v}{\Delta t}\rightarrow \Delta t=\frac{v_i}{g}$

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