The wind blows a 0.50 kg Styrofoam box, initially at rest, across level ground and then up an incline where it comes to a rest. The box slides 6.0 m on level ground before sliding up the 20.0
Ovincline. If the effective coefficient of friction between the ground and the box is 0.60, and the wind exerts a constant, horizontally-directed force of 5.0 N, how far up the incline will the box be blown by this wind before it
comes to a stop?
Horizontal motion
m •a=F-μ•m•g.
a= (F-μ•m•g )/m=
=F/m - μ•g =5/0.5 – 0.6•9.8 = 4.12 m/s²
s1=at²/2 => t=sqrt(2•s/a) =1.7 s.
v=at=4.12•1.7 = 7.03 m/s.
The law of conservation of energy for incline:
mv²/2 +W(F) = W(fr) +ΔPE.
mv²/2 +F•x•cosθ = μ•m•g•cosθ•x+m•g•x•sinθ.
x= mv²/2{μ•m•g•cosθ-F•cosθ - m•g•sinθ} = 0.5 m.
To solve this problem, we need to calculate the net force acting on the Styrofoam box at each stage of its motion.
First, let's consider the box sliding on level ground. The only forces acting on the box are the force of friction and the force exerted by the wind. The weight of the box can be ignored since it cancels out with the normal force from the ground. The net force can be calculated as follows:
friction force = coefficient of friction * normal force
= 0.60 * (mass of the box * acceleration due to gravity)
= 0.60 * (0.50 kg * 9.8 m/s^2)
= 2.94 N
Since the box is initially at rest, the net force is only due to the wind force:
net force on the box on level ground = force of wind = 5.0 N
Since the force of friction is less than the force of wind, there is a resulting net force that accelerates the box on level ground. We can use Newton's second law to determine the acceleration of the box:
net force = mass of the box * acceleration
5.0 N = 0.50 kg * acceleration
acceleration = 10 m/s^2
Using the kinematic equation, we can determine the time it takes for the box to slide 6.0 m:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
6.0 m = 0 + (0.5 * 10 m/s^2 * time^2)
3.0 = 5 * time^2
time^2 = 3.0/5
time = √(3.0/5) ≈ 0.77 s
Now, let's consider the box sliding up the incline. The only force acting on the box is the friction force, which opposes the motion. The net force on the box can be calculated as follows:
friction force = coefficient of friction * normal force
= 0.60 * (mass of the box * acceleration due to gravity * cosθ)
= 0.60 * (0.50 kg * 9.8 m/s^2 * cos20°)
= 2.34 N
Since the box is coming to a stop, the net force on the box up the incline is equal to zero. We can set up the equation:
net force on the box up the incline = force of wind - friction force = 0
5.0 N - 2.34 N = 0
2.66 N = 0
Since the net force is zero, there is no resulting acceleration. Therefore, the box will stop immediately after reaching the incline. It will not slide up the incline any further.