The problem propmpts us with the info:
"Two friends play with a card-board box on a grassy hill side.
The boy in the box has a mass of 50.0 kg and the box has a mass of 5.00 kg.
The incline of the hill is 44° and its height is 10 meter.
Assume that friction plays a negligible role on the hill.
You will need to know the length of the incline."
We need to find the acceleration, velocity at the bottom of the hill, and the friction force in Newtons (this is when friction is no longer negligible. "The box slides horizontally across the grass at the bottom of the incline, coming to a stop 13.1 meters from the base of the incline. Assume friction plays a significant role bringing the box to a stop").
I have the Max=mgsine44-uN, May=n-49cos44=0 by rotating the motion map so acceleration is horizontal. I found n to be equal to 49cos44, 35.2. When I plugged in "n" to the max equation I found u to equal .404. It says friction is negligible, so does that mean u=0? But, if u=0 then you can't find acceleration through a=ug. I used trig to find the length of the incline to be 14.4 m. To find velocity at the bottom of the hill, I made an xvvat chart and used the values initial velocity=0, my found acceleration (which apparently is wrong), and the position, x, as 14.4. My friction should be f=un, but my n must not be correct.
To solve this problem, let's break it down step by step:
1. Calculate the gravitational force acting on the boy and the box:
The gravitational force (Fg) can be calculated using the mass (m) and the acceleration due to gravity (g).
For the boy: Fg_boy = m_boy * g = 50.0 kg * 9.8 m/s^2 = 490 N
For the box: Fg_box = m_box * g = 5.00 kg * 9.8 m/s^2 = 49 N
2. Resolve the gravitational force into components:
The gravitational force can be resolved into two components - one parallel to the incline and one perpendicular to the incline.
For the boy: Fg_boy_parallel = Fg_boy * sin(angle) = 490 N * sin(44°) ≈ 333.7 N
For the boy: Fg_boy_perpendicular = Fg_boy * cos(angle) = 490 N * cos(44°) ≈ 372.2 N
For the box: Fg_box_parallel = Fg_box * sin(angle) = 49 N * sin(44°) ≈ 33.5 N
For the box: Fg_box_perpendicular = Fg_box * cos(angle) = 49 N * cos(44°) ≈ 43.8 N
3. Calculate the normal force acting on the box:
The normal force (N) is the force exerted by the incline perpendicular to its surface.
Since the boy stands on the incline, the normal force acting on the boy is equal to Fg_boy_perpendicular.
For the box: N = Fg_box_perpendicular = 43.8 N
4. Determine if friction is negligible:
The problem states that friction plays a negligible role on the hill, so we assume the coefficient of friction (u) is zero.
5. Calculate the acceleration of the box:
Since the friction is negligible and there are no other horizontal forces acting on the box, the net force in the horizontal direction is equal to the force parallel to the incline acting on the box.
The force parallel to the incline (F_parallel) can be calculated as: F_parallel = Fg_box_parallel = 33.5 N
Using Newton's second law (F = m * a), we can solve for the acceleration (a):
33.5 N = (m_box + m_boy) * a
a = 33.5 N / (5.00 kg + 50.0 kg) ≈ 0.621 m/s^2
6. Calculate the velocity at the bottom of the hill:
The distance traveled by the box along the incline (s) is given as 13.1 meters.
Using the equation of motion (v^2 = u^2 + 2 * a * s), where u is the initial velocity, we can solve for v:
v^2 = 0^2 + 2 * 0.621 m/s^2 * 13.1 m
v ≈ 5.76 m/s
7. Friction force when the box stops:
When the box comes to a stop at the bottom of the incline, the friction force is responsible for bringing it to rest.
The friction force (F_friction) can be calculated using the equation F_friction = u * N, where u is the coefficient of friction and N is the normal force.
Since the question states that friction plays a significant role now, we assume a non-zero coefficient of friction.
However, since the problem does not provide the coefficient of friction, it cannot be determined without additional information.
So, to summarize, you correctly found the acceleration to be approximately 0.621 m/s^2. The velocity at the bottom of the hill is approximately 5.76 m/s. However, without knowing the coefficient of friction, it is not possible to calculate the friction force.