A child with a mass of 83 kg is on top of a water slide that is 12 m long and inclined 30 degrees with the horizontal. How fast is the child going at the bottom of the slide (a) if the slide is frictionless? (b) if the coefficient of friction is 0.30?
a. h = 12*s1n30 = 6 m.
V^2 = Vo + 2g*h = 0 + 19.6*6 = 117.6
V = 10.84 m/s.
b. m*g=83kg * 9.8N/kg = 813.4 N.=Wt. of
the child.
Fn = mg*cos30 = 813.4*cos30 = 704.4 N. = Normal = Force perpendicular to the
slide.
Fk = u*Fn = 0.30 * 704.4 = 211.3 N. =
Force of kinetic friction.
PEmax = mg*h-Fk*L = 813.4*6-211.3*12 =
2344.5 Joules.
KE + PE = 2344.5
KE + 0 = 2344.5 At bottom of slide.
KE = 0.5m*V^2 = 2344.5
41.5V^2 = 2344.5
V^2 = 56.5
V = 7.52 m/s.
Well, hold your water! Let's figure this out one splash at a time.
(a) If the slide is frictionless, then the only forces acting on the child are gravity and normal force. We can break the gravitational force into two components: one parallel to the slide and one perpendicular. The perpendicular component doesn't affect the child's speed down the slide, so we can focus on the parallel component, which can be found using the equation:
Force parallel = mg * sin(theta)
where m is the mass (83 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and theta is the angle of the slide (30 degrees).
Force parallel = 83 kg * 9.8 m/s^2 * sin(30 degrees)
Solving this equation gives us the force parallel to be approximately 405.6 N.
Using Newton's second law (F = ma), we can find the child's acceleration:
Acceleration = force parallel / mass
Acceleration = 405.6 N / 83 kg
Acceleration is roughly 4.89 m/s^2.
Now, we can use a kinematic equation to find the child's velocity at the bottom of the slide, assuming they start from rest:
v^2 = u^2 + 2as
Since the child starts from rest, the initial velocity (u) is 0. So the equation simplifies to:
v^2 = 2as
Plugging in the values, we get:
v^2 = 2 * 4.89 m/s^2 * 12 m
Solving this equation gives us the child's velocity at the bottom of the frictionless slide to be about 17.32 m/s.
(b) Now let's slide into the case where the coefficient of friction is 0.30. The frictional force can be calculated using the equation:
Force friction = coefficient of friction * normal force
To find the normal force, we have to use some trigonometry. The normal force can be found by:
Force normal = mg * cos(theta)
Force normal = 83 kg * 9.8 m/s^2 * cos(30 degrees)
Solving this equation gives us the normal force to be approximately 706.14 N.
Now, we can find the force friction:
Force friction = 0.30 * 706.14 N
Force friction is about 211.84 N.
Since the slide is inclined, the child will start off with a higher velocity compared to the frictionless case. So we need to consider the initial velocity (u) in the kinematic equation:
v^2 = u^2 + 2as
Now, solving for the child's velocity at the bottom of the slide:
v^2 = u^2 + 2as
v^2 = 0 + 2 * (405.6 N - 211.84 N) / 83 kg * 12 m
Solving this equation gives us the child's velocity at the bottom of the slide, with a coefficient of friction of 0.30, to be approximately 7.87 m/s.
So there you have it! The child will be sliding at about 17.32 m/s when the slide is frictionless, and about 7.87 m/s when there is a coefficient of friction of 0.30. Don't slip on a banana peel while trying to calculate those speeds!
To solve this problem, we can use the principles of physics, specifically Newton's second law and the equations of motion.
(a) If the slide is frictionless, the only force acting on the child is the force of gravity. We can use the equation:
m * g * sin(theta) = m * a
where:
m = mass of the child = 83 kg
g = acceleration due to gravity = 9.8 m/s^2
theta = angle of inclination = 30 degrees
a = acceleration of the child
Rearranging the equation, we get:
a = g * sin(theta)
Substituting the given values into the equation:
a = 9.8 m/s^2 * sin(30 degrees)
Calculating the value:
a = 4.9 m/s^2 * 0.5
a = 2.45 m/s^2
Since the child is at rest initially, the initial velocity (u) is zero. We can use the equation of motion:
v^2 = u^2 + 2 * a * s
where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
Since the child is sliding down the slide, the distance (s) is the length of the slide, which is 12 m. Substituting the values into the equation:
v^2 = 0 + 2 * 2.45 m/s^2 * 12 m
Calculating the value:
v^2 = 2.45 m/s^2 * 24 m
v^2 = 58.80 m^2/s^2
Taking the square root of both sides:
v = √58.80 m/s
Calculating the value:
v ≈ 7.67 m/s
Therefore, the child is going approximately 7.67 m/s at the bottom of the slide if the slide is frictionless.
(b) If the coefficient of friction is 0.30, we need to consider the force of friction in addition to the force of gravity.
The force of friction can be calculated using the equation:
f = μ * m * g
where:
μ = coefficient of friction = 0.30
Substituting the values:
f = 0.30 * 83 kg * 9.8 m/s^2
Calculating the value:
f = 0.30 * 813.4 N
f ≈ 244.02 N
The net force acting on the child can be calculated using the equation:
net force = m * a
where:
m = mass of the child = 83 kg
a = acceleration of the child
Since the child is moving downhill, the net force is the force of gravity minus the force of friction:
net force = m * g * sin(theta) - f
Substituting the values:
net force = 83 kg * 9.8 m/s^2 * sin(30 degrees) - 244.02 N
Calculating the value:
net force = 83 kg * 4.9 m/s^2 - 244.02 N
net force = 406.70 N - 244.02 N
net force = 162.68 N
Finally, we can use the net force to calculate the acceleration:
net force = m * a
162.68 N = 83 kg * a
Solving for a:
a = 162.68 N / 83 kg
Calculating the value:
a ≈ 1.96 m/s^2
Using the equation of motion:
v^2 = u^2 + 2 * a * s
where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
Since the child is sliding down the slide, the distance (s) is still 12 m. Substituting the values into the equation:
v^2 = 0 + 2 * 1.96 m/s^2 * 12 m
Calculating the value:
v^2 = 1.96 m/s^2 * 24 m
v^2 = 47.04 m^2/s^2
Taking the square root of both sides:
v = √47.04 m/s
Calculating the value:
v ≈ 6.86 m/s
Therefore, the child is going approximately 6.86 m/s at the bottom of the slide if the coefficient of friction is 0.30.
To determine the speed of the child at the bottom of the water slide, we can first find the gravitational potential energy at the top of the slide and then use the principle of conservation of mechanical energy.
To find the gravitational potential energy at the top of the slide, we can use the formula:
PE = m * g * h,
where m is the mass of the child (83 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the slide (12 m).
(a) If the slide is frictionless, the total mechanical energy at the top of the slide is equal to the gravitational potential energy. Therefore, the speed of the child at the bottom of the slide can be found using the formula:
m * g * h = (1/2) * m * v^2,
where v is the final velocity of the child at the bottom of the slide.
Simplifying the equation, we get:
v^2 = 2 * g * h,
v = sqrt(2 * g * h).
Substituting the values, we get:
v = sqrt(2 * 9.8 * 12) = sqrt(235.2) ≈ 15.34 m/s.
(b) If the coefficient of friction on the slide is given (0.30), we need to consider the work done against friction. The work done against friction can be given by the formula:
Work = friction * distance,
where the friction is given by:
friction = coefficient of friction * normal force,
and the normal force is equal to the weight of the child, which is given by:
normal force = m * g.
The work done against friction can be calculated by multiplying the friction force by the distance along the slide. In this case, the distance is equal to the length of the slide (12 m). Therefore, the work done against friction is:
Work = (coefficient of friction * m * g) * distance,
Work = (0.30 * 83 * 9.8) * 12 = 293.04 J.
Now, we can find the speed at the bottom of the slide by considering the conservation of mechanical energy:
(m * g * h) - (Work) = (1/2) * m * v^2.
Substituting the values, we get:
(83 * 9.8 * 12) - 293.04 = (1/2) * 83 * v^2.
Simplifying the equation, we get:
v^2 = ((83 * 9.8 * 12) - 293.04) / (1/2 * 83),
v^2 ≈ 1331.13,
v ≈ sqrt(1331.13) ≈ 36.48 m/s.
Therefore, the speed of the child at the bottom of the slide is approximately 36.48 m/s if the coefficient of friction is 0.30.