The spring in the figure has a spring
constant of 1000 N/
m. After being
compressed by 15 cm it launches a 200 g
block out onto a frictionless horizontal
surface. The block then moves up an
incline, where the coefficient of friction
is 0.20, eventually becoming airborne as
it reaches a plateau 2.0 m higher th
an the original surface.
Determine what distance
d
the block sails through the air
To determine the distance the block sails through the air, we can use the principle of conservation of mechanical energy.
First, let's calculate the potential energy stored in the compressed spring.
The potential energy of a spring can be given by the formula:
Potential energy (PE) = (1/2) * k * x^2
Where:
k = spring constant = 1000 N/m
x = compression of the spring = 15 cm = 0.15 m
PE = (1/2) * 1000 * (0.15)^2
PE = 11.25 J
This potential energy is converted into kinetic energy as the block is launched into the air.
Next, let's calculate the speed of the block when it leaves the incline.
The mechanical energy at the base of the incline is equal to the mechanical energy at the top of the incline:
KE_top + PE_top = KE_base + PE_base
At the top of the incline, the block has only kinetic energy:
KE_top = (1/2) * m * v^2
Where:
m = mass of the block = 200 g = 0.2 kg
v = speed of the block at the top of the incline (to be determined)
KE_top = (1/2) * 0.2 * v^2
At the base of the incline, the block has kinetic energy and potential energy due to its height above the initial surface:
KE_base = (1/2) * m * v_base^2
PE_base = m * g * h
Where:
v_base = speed of the block at the base of the incline (to be determined)
g = acceleration due to gravity = 9.8 m/s^2
h = height of the plateau above the initial surface = 2.0 m
KE_base = (1/2) * 0.2 * v_base^2
PE_base = 0.2 * 9.8 * 2.0
KE_base = PE_base
(1/2) * 0.2 * v_base^2 = 0.2 * 9.8 * 2.0
Solving for v_base:
v_base^2 = 9.8 * 2 * 2
v_base^2 = 39.2
v_base = √39.2
v_base = 6.26 m/s
Now, let's find the speed of the block at the top of the incline:
KE_top = (1/2) * 0.2 * v^2
11.25 = (1/2) * 0.2 * v^2
Solving for v:
0.4 * v^2 = 11.25
v^2 = 11.25 / 0.4
v^2 = 28.125
v = √28.125
v = 5.31 m/s
Finally, we can determine the distance the block sails through the air using the formula:
Distance (d) = (v^2) / (2 * g * μ)
Where:
μ = coefficient of friction = 0.20
d = (5.31^2) / (2 * 9.8 * 0.20)
d = 2.80 m
Therefore, the block sails through the air for a distance of 2.80 meters.