A 1.8 tonne railway truck moving at a speed of 3m/s up a steady incline comes to a stop after 45m. If the sum of the friction opposing the truck amounts to 80N What is the angle of elevation of the slope?

I found Fg using m*g
THen I tried using v^2=u^2 + 2as to find the acceleration so I could find Fnet and then use that to find the Fapplied
Adding that to the friction of 80N (because it is in the same direction).
Then I thought to use trigonometry to solve for the angle only it didn'tt give me the correct answer...

Please could you help???

Wt. = 1.8 Tonne.

Vo = 3 m/s.
d = 45 m.
Fk = 80 N.

Mass = 1.8 * 1000kg = 1800 kg.

Fg = m*g = 1800 * 9.8 = 17,640 N.

KE + PE = mg*h-Fk*d
KE + PE = mg*h-80*45
0.5m*Vo^2 + 0 = mg*h-3600
900*9 + 3600 = 17640h
17,640h = 11,700
h = 0.6633 m.

sinA = h/d = 0.6633/45 = 0.01474
A = 0.84o = Angle of elevation.

WHAT DOES A DISPLACEMENT-TIME BALLON TYPE GRAPH REPRESENTS?

Of course, I'd be happy to help you with this problem! Let's break it down step by step.

Step 1: Calculate the weight of the truck:
To determine the weight (force) of the truck, we need to use the formula `Fg = m * g`, where `m` is the mass and `g` is the acceleration due to gravity. Since the truck weighs 1.8 tonnes, we need to convert it to kilograms first. Recall that 1 tonne is equal to 1000 kilograms. Therefore, the mass of the truck is 1.8 tonnes * 1000 kg/tonne = 1800 kg.

Now we can find the weight of the truck: `Fg = 1800 kg * 9.8 m/s^2 = 17640 N`.

Step 2: Determine the net force acting on the truck:
To calculate the net force, we'll use the equation `Fnet = Fapplied - Ffriction`, where `Fapplied` is the applied force and `Ffriction` is the force of friction opposing the truck's motion. In this case, the truck is moving up the incline, so the applied force is acting against gravity. Since the truck comes to a stop, the net force must be zero.

Therefore, `Fapplied - Ffriction = 0`, which means `Fapplied = Ffriction = 80 N`.

Step 3: Calculate the acceleration:
We can use the equation `v^2 = u^2 + 2as`, where `v` is the final velocity (0 m/s) since the truck comes to a stop, `u` is the initial velocity (3 m/s), `a` is the acceleration, and `s` is the displacement. Rearranging the equation, we have `a = (v^2 - u^2) / (2s) = (0^2 - 3^2) / (2 * 45) = -9/30 = -0.3 m/s^2`. The negative sign indicates that the acceleration is opposing the motion.

Step 4: Determine the net force due to the incline:
To find the net force acting on the truck due to the incline, we'll use the equation `Fnet = m * g * sin(theta)`, where `m` is the mass of the truck, `g` is the acceleration due to gravity, and `theta` is the angle of elevation of the slope. Recall that there is no motion along the vertical direction (perpendicular to the incline), so the net force in that direction is zero.

Therefore, `Fnet = m * g * sin(theta) = 17640 N * sin(theta) = 0`.

Since the net force is zero, `sin(theta) = 0`, which means `theta = sin^(-1)(0) = 0°`. This indicates that the slope is horizontal or flat.

So, according to the given information, the angle of elevation of the slope is 0°, meaning it is flat.

I hope this explanation helps! Let me know if you have any further questions.