A 37 kg child roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 17 degrees with the horizontal. If there's no friction, what is the kinetic energy of the child at the bottom of the incline?
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To find the kinetic energy of the child at the bottom of the incline, we need to use the formula for kinetic energy: KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
First, let's find the velocity of the child at the bottom of the incline. Since the child is rolling down an incline at an angle, we need to consider the component of the gravitational force that is parallel to the incline. This force can be calculated using the formula: F = m * g * sin(theta), where F is the force, m is the mass, g is the acceleration due to gravity, and theta is the angle of the incline.
In this case, the force parallel to the incline is equal to the weight of the child, so we can set the two equations equal to each other: m * g * sin(theta) = m * a, where a is the acceleration.
By canceling out the mass, we can solve for the acceleration: a = g * sin(theta).
Now that we have the acceleration, we can find the velocity using the equation: vf^2 = vi^2 + 2 * a * d, where vf is the final velocity, vi is the initial velocity (which is zero since the child is initially at rest), a is the acceleration, and d is the distance traveled down the incline.
Plugging in the values: vf^2 = 0^2 + 2 * (g * sin(theta)) * d.
To simplify, we can substitute g * sin(theta) with a: vf^2 = 2 * a * d.
Now we can calculate vf: vf = sqrt(2 * a * d).
Finally, we can calculate the kinetic energy using the formula KE = 1/2 * m * vf^2: KE = 1/2 * m * (sqrt(2 * a * d))^2.
Plugging in the values: KE = 1/2 * 37 kg * (sqrt(2 * (9.8 m/s^2) * (sin(17 degrees)) * 2m))^2.
Simplifying, we find that the kinetic energy of the child at the bottom of the incline is approximately equal to 946 J.