A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.21 m/s at the bottom of the incline. Assume
g = 9.80 m/s2.
How far up the incline (measured parallel to the surface of the incline) does the block travel before it starts to slide back down?
F=ma=mgsin(theta)
is used to find the acceleration
mass does not matter, as it cancels out
v^2=v0^2 +2ax
plug in:
initial velocity is given as 4.21 m/s,
final velocity is 0 m/s, as it changes direction,
and the acceleration found earlier to solve for x
To determine how far up the incline the block travels before it starts to slide back down, we can use the principle of conservation of energy.
At the bottom of the incline, the block possesses both kinetic energy and gravitational potential energy. As it moves up the incline, its kinetic energy is converted to gravitational potential energy.
The kinetic energy (K) of an object is given by the equation K = (1/2)mv^2, where m is the mass of the object and v is its velocity.
The gravitational potential energy (U) of an object at a height h is given by the equation U = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
Since the incline is frictionless, there is no work done against friction. Therefore, the total mechanical energy (E) of the block will be conserved, and we can equate the initial and final energies.
At the bottom of the incline, all of the energy is kinetic energy, so E = K = (1/2)mv^2.
As the block moves up the incline, it eventually reaches a height where its velocity becomes zero before it starts sliding back down. At this point, all of the energy is gravitational potential energy, so E = U = mgh.
Setting these two equations equal to each other, we have:
(1/2)mv^2 = mgh.
The mass of the block cancels out, leaving us with:
(1/2)v^2 = gh.
Since we know the angle of the incline (30.0°) and the acceleration due to gravity (g = 9.80 m/s^2), we can find the height (h) as follows:
h = (1/2)v^2 / g.
Plugging in the given values for v and g:
h = (1/2)(4.21 m/s)^2 / (9.80 m/s^2).
Calculating this expression:
h = 0.091 meters.
Therefore, the block travels approximately 0.091 meters up the incline before it starts to slide back down.