A block is given an initial velocity of 5.80 m/s up a frictionless 19.5° incline. How far up the incline does the block slide before coming to rest?
KE= PE
m•v²/2=m•g•h =m•g•s•sinα,
s= v²/2•g•sinα
how far up theincline does the bliok silde before coming to rest?
To find how far up the incline the block slides before coming to rest, we can use the principles of projectile motion and the work-energy theorem.
First, let's break down the initial velocity into its component vectors. The vertical component is given by V_y = V * sin(θ), where V is the magnitude of the initial velocity (5.80 m/s) and θ is the angle of the incline (19.5°). So, V_y = 5.80 m/s * sin(19.5°).
Next, we use the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy. When the block comes to rest, its kinetic energy becomes zero. The initial kinetic energy is given by KE = (1/2) * m * V^2, where m is the mass of the block. Since the incline is frictionless, there is no work done against friction.
Equating the initial kinetic energy to zero, we have:
(1/2) * m * V^2 = 0
Simplifying the equation, we get:
V^2 = 0
Thus, V = 0, meaning the block comes to rest.
Now, let's look at the vertical motion of the block. The vertical distance covered by the block before coming to rest is given by the equation:
d = (V_y^2) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values, we have:
d = (5.80 m/s * sin(19.5°))^2 / (2 * 9.8 m/s^2)
Solving this equation will give us the distance up the incline that the block slides before coming to rest.
Note: Be sure to convert the angle from degrees to radians when using the sin function in your calculations.