How far does a 5-kg block with an initial speed of 3 m/s slide up a frictionless 28° incline plane?
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To find how far the 5-kg block slides up the frictionless 28° incline plane, we can use the principles of physics and trigonometry.
First, we need to determine the force acting in the direction of the incline. This force is equal to the component of the weight of the block that is parallel to the incline. We can calculate this using the formula:
Force_parallel = mg * sin(θ)
Where:
m = mass of the block = 5 kg
g = acceleration due to gravity ≈ 9.8 m/s²
θ = angle of the incline = 28°
Next, we can calculate the acceleration of the block using Newton's second law of motion:
Force_parallel = mass * acceleration
Rearranging the equation, we have:
acceleration = Force_parallel / mass
Substituting the values we know:
acceleration = (mg * sin(θ)) / mass
Now, we can use the equations of motion to determine how far the block slides up the incline.
The equation to calculate the distance travelled by an object under constant acceleration is:
distance = (initial velocity²) / (2 * acceleration)
We are given the initial velocity of the block as 3 m/s, and we calculated the acceleration in the previous step.
Substituting the values into the equation:
distance = (3²) / (2 * acceleration)
Finally, plug in the calculated value for acceleration:
distance = (3²) / (2 * ((mg * sin(θ)) / mass))
Simplifying the equation further by canceling out the mass:
distance = (3² * mass) / (2 * mg * sin(θ))
Plugging in the known values:
distance = (3² * 5) / (2 * 5 * 9.8 * sin(28°))
Evaluating the expression:
distance ≈ 1.15 meters
Therefore, the 5-kg block with an initial speed of 3 m/s slides approximately 1.15 meters up the frictionless 28° incline plane.
It slides until the. Initial kinetic energy equals the gain in potential energy. If S is the distance travelled up the ramp,
(M/2)*Vo^2 = M g S sin28
S = Vo^2/(2 g sin28)