A 20 kg child on roller skates, initially at rest, rolls 4 m down an incline at an angle of 30 degrees with the horizontal. If there is no friction between incline and skates, what is the kenetic energy of the child at the bottom of the incline? (g= 9.81 m/s^2)
There are two ways of doing this problem, the energy way and the work way.
The energy way is to say the potential energy lost by moving down is the kinetic energy gained at the bottom.
Potential energy lost = m g times the change in height
The change in height is from:
sin (30) = change in height / 4
so change in height = 4 sin 30 = 2 meters down
so the potential energy lost is:
m * 9.8 * 2 which is 19.6 m Joules
That 19.6 m Joules is the kinetic energy at the bottom
KE = 19.6 m
since m = 20 kg
KE = 20*19.6 = 392 Joules
NOW the other way
The work done is the force down the ramp times the distance run
The component of force down the ramp is
m g cos 60 = 20 *9.8 * .5 = 98 Newtons
That 98 newtons pushes a distance of 4 meters so the work done is
98 * 4 = 392 Joules
That work done is used to increase the kinetic energy of the child so the kinetic energy is also 392 Joules.
Well, let's calculate it!
The potential energy of the child at the top of the incline can be calculated using the formula:
PE = m * g * h
where m is the mass of the child, g is the acceleration due to gravity, and h is the vertical height of the incline.
Since the incline is at an angle of 30 degrees with the horizontal, the vertical height can be calculated using trigonometry:
h = 4 * sin(30)
So, h = 2 meters.
Now, let's plug in the values into the formula:
PE = 20 * 9.81 * 2
PE = 392.4 Joules
Since there is no friction, the potential energy will be completely converted into kinetic energy at the bottom of the incline. Therefore, the kinetic energy of the child at the bottom of the incline will be equal to 392.4 Joules.
And hey, look at that! The child gained some serious energy rolling down that incline. Time to put all that potential into motion!
To calculate the kinetic energy of the child at the bottom of the incline, we can use the formula:
Kinetic Energy (KE) = 1/2 * mass * velocity^2
First, we need to find the velocity of the child at the bottom of the incline. We can use the principle of conservation of mechanical energy, assuming no friction.
The initial potential energy (PE) of the child at the top of the incline is given by:
PE = mass * gravity * height
where mass = 20 kg, gravity (g) = 9.81 m/s^2, and height = 4 m.
PE = 20 kg * 9.81 m/s^2 * 4 m = 784.8 J
Since there is no friction, this potential energy is converted entirely into kinetic energy at the bottom of the incline, so KE = PE.
Therefore, the kinetic energy of the child at the bottom of the incline is 784.8 J.
To find the kinetic energy of the child at the bottom of the incline, we need to first calculate the child's final velocity at the bottom using the laws of motion.
The child is moving in a 2-dimensional plane, with two forces acting on them: the force of gravity pulling downwards and the force due to the incline. The force due to gravity can be split into two components: one parallel to the incline (mg sinθ, where m is the mass of the child and θ is the angle of the incline) and one perpendicular to the incline (mg cosθ).
The force due to the incline can be calculated using the equation: F = ma, where F is the force, m is the mass, and a is the acceleration along the incline. Since the child is initially at rest, the initial velocity (u) is 0.
Now, let's calculate the gravitational force components and the acceleration along the incline:
Force parallel to the incline = mg sinθ
Force perpendicular to the incline = mg cosθ
Acceleration along the incline = F / m = (mg sinθ) / m = g sinθ
Since the incline is at an angle of 30 degrees, the acceleration along the incline can be calculated as:
Acceleration along the incline = g sin(30°)
Next, let's calculate the distance traveled by the child down the incline:
Distance (s) = 4 m
Using the equation of motion: s = ut + 0.5at², where s is the distance, u is the initial velocity, a is the acceleration, and t is the time taken:
4 = 0.5 * (g sin(30°)) * t²
Since the child starts from rest, the initial velocity (u) is 0. Rearranging the equation, we get:
t = √(8s / (g sin(30°)))
Now, let's calculate the time taken.
t = √(8 * 4 / (9.81 * sin(30°)))
Simplifying the equation:
t ≈ 1.168 seconds
Now that we know the time taken, we can calculate the final velocity (v) using the equation: v = u + at. Since the initial velocity is 0, the equation simplifies to: v = at.
v = (g sin(30°)) * t
Plugging in the values:
v ≈ (9.81 * sin(30°)) * 1.168
v ≈ 5.087 m/s
Finally, to calculate the kinetic energy (KE) of the child at the bottom of the incline, we can use the equation: KE = 0.5 * m * v².
Plugging in the values:
KE = 0.5 * 20 * (5.087)²
KE ≈ 517.84 Joules
Therefore, the kinetic energy of the child at the bottom of the incline is approximately 517.84 Joules.