In the case of an object on an incline, the normal force is mgcos in the normal direction, while the combined force of gravity + normal force is mgsin down the incline. A proof of this will be provided at the end.
In the case of two surfaces sliding against each other (as opposed to a collision), normal force is a force that always acts normal to said surfaces, and is always just enough to prevent them from passing through each other, when all other forces are factored in. Kinetic friction, meanwhile, always acts parallel to the surfaces that are sliding against each other, and thus cannot push two surfaces closer together or farther apart. As a result, it does not need to be factored in when calculating normal force, and the initial statement still holds.
Without friction, the total force is mgsin(35°) down the incline.
The force of friction will apply in the direction opposite the velocity, and with a magnitude equal to the normal force times μ_k. Since the normal force is mgcos, and the kinetic friction is 0.1, that means, as long as Tromp is moving up the incline and not down, the force of friction is 0.1mgcos down the incline.
We therefore have a net force of mg*(sin(35°)+0.1cos(35°)) down the incline.
For the sake of clarity, I will not be abbreviating meters or second, so as to avoid confusing m (mass) with m (meter).
1.)
F = ma, so a = F/m
Tromp's acceleration is therefore g*(sin(35°)+0.1cos(35°)) = 9.8*(0.655) = 6.42 meters/second²
Tromp's acceleration is 6.42 meters/second² down the ramp
2.)
Tromp's initial speed is 15 meters/second.
Calculation assuming you were taught the concept of work/energy:
Tromp's initial kinetic energy is mv²/2 = m*15²/2 = 112.5m. We're keeping m as a variable so we can reuse this for Nedib in part 3.
Work is the change in kinetic energy, and when a constant force is applied, the work is force * displacement * cos(angle between the two). The net displacement is up the ramp, while the net force applies down the ramp, therefore the angle between the two is 180°, and the cosine is -1.
force = 6.42 * mass
displacement = length of hypotenuse of ramp = 3 / sin(35°) = 5.23
work = 6.42*mass * 5.23 * -1 = -33.6*mass
So, at the top of the ramp, the kinetic energy has gone down by 33.6*mass, and we have 78.9*mass left.
KE = mv²/2
v² = KE*2/mass = 78.9*mass * 2/mass = 157.8
v = √(157.8) = 12.6 meters/second.
Tromp's velocity at the top of the ramp is 12.6 meters/second up the ramp.
Calculation assuming you weren't taught that concept:
Equation of motion = a*t²/2 + v_0*t + pos_0
Equation of motion = -6.42*t²/2 + 15*t + 0 = -3.21*t² + 15*t
Find t at the top of the ramp: The length of the ramp is the hypotenuse, which is 3/sin(35°) = 5.23
-3.21t² + 15t = 5.23
-3.21t² + 15t - 5.23 = 0
By the quadratic equation, we have t = (-15±√(15²-4*-3.21*-5.23))/(2*-3.21) = 0.379 or 4.29.
Since both times are positive, we can assume the first one would be Tromp going up the ramp, and the second time would happen after Tromp's velocity goes to 0 (though that actually wouldn't happen, even if the ramp were long enough, as Tromp would either stop due to static friction when his velocity goes to 0, or he would fall back downward, but his kinetic friction would flip directions, thus changing the equation of motion). In any case, this means that 0.379 is the correct time.
Equation of velocity: a*t + v_0 = -6.42t + 15
Velocity at the top = -6.42 * 0.379 + 15 = 12.6 meters/second.
3.)
Our equations were completely unaffected by Tromp's mass. That is to say, if we were to perform the same calculations, but under the assumption that the mass is 65kg and not 70kg, we'd get the exact same result.
Therefore, Nedib will have the exact same acceleration as Tromp (6.42 meters/second² down the ramp), and will have the exact same velocity at the top of the ramp (12.6 meters/second up the ramp).
Proof of initial statement:
If we assume the object is going to the right (positive x), and that upwards is positive y, then we have:
Gravity = <0,-mg>
Normal Force = N*<-sin(θ), cos(θ)>
Net force: <-Nsin(θ), Ncos(θ)-mg>
Net force in the direction of the normal: <-Nsin(θ), Ncos(θ)-mg> ⋅ <-sin(θ), cos(θ)> = Nsin(θ)² + Ncos(θ)² - mgcos(θ) = N - mgcos(θ).
(Note ⋅ represents the dot product, which measures how parallel two vectors are times how long they are. In this case, we're using it to find the force when measured in the direction <-sin(35°), cos(35°)>)
So N is the value such that that directional force is 0. That is to say, N - mgcos(θ) = 0, or that N = mgcos(θ).
It should be noted that this applies both to upward inclines and downward inclines. To do the same calculation on a downward incline, do all the above, but with sin(θ) replaced with sin(-θ), or -sin(θ). You will get the same result.
