I need help with this physics problem: A child slides down a playground slide; the coefficient of kinetic friction is uk=0.25, and angle that the slide makes with the horizontal is 28.0 degree. She begins from rest and slides through a vertical height of 3.5 m. With what speed does she exit the slide?
There are several ways to do this. Let's try conservation of energy. The loss of potential coming down (M g H)is equal to the gain in kinetic energy, (1/2) M V^2, PLUS work done against friction while sliding. The length of the slide is H/sin 28 = 7.46m, where H = 3.5 m is the vertical height. The friction force is M g cos28*uk. The friction work done is therefore
W = M g (H/sin 28)* cos28 *uk
= M g uk/tan28
M g H = M g H uk/tan 28 + (1/2)M V^2
Note that the M's cancel. That's good, because they did not tell you what the mass was.
V^2 = 2 g H [1 - uk/tan28]
To find the speed with which the child exits the slide, we can use the conservation of mechanical energy. The initial potential energy is converted to kinetic energy at the bottom of the slide.
1. Determine the gravitational potential energy:
The potential energy is given by PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the vertical height.
The mass of the child is not given, but it cancels out when calculating the velocity, so we don't need it.
PE = mgh = gh
2. Determine the kinetic energy:
The kinetic energy is given by KE = 0.5mv^2, where m is the mass, and v is the velocity.
We'll solve this equation for v.
3. Apply conservation of mechanical energy:
The total mechanical energy at the start (top of the slide) is equal to the total mechanical energy at the end (bottom of the slide).
Initial mechanical energy (E_initial) = PE
Final mechanical energy (E_final) = KE + work done against friction
The work done against friction is given by the equation: work = μk * mg * d, where μk is the coefficient of kinetic friction, m is the mass, g is the acceleration due to gravity, and d is the displacement.
At the start, all the potential energy is converted to kinetic energy and work against friction:
E_initial = E_final
gh = 0.5mv^2 + μk * mg * d
Substituting the given values:
g = 9.8 m/s^2
h = 3.5 m
μk = 0.25
d = 3.5 m
9.8 * 3.5 = 0.5 * v^2 + 0.25 * 9.8 * 3.5
Solve for v by rearranging the equation:
v^2 = (2 * (9.8 * 3.5 - 0.25 * 9.8 * 3.5))/0.5
v^2 = 2 * (9.8 * 3.5 - 0.25 * 9.8 * 3.5)
v^2 = 2 * 9.8 * 3.5 * (1 - 0.25)
v^2 = 2 * 9.8 * 3.5 * 0.75
v^2 = 64.365
Taking the square root of both sides:
v ≈ 8.03 m/s
Therefore, the child exits the slide with a speed of approximately 8.03 m/s.
To find the speed at which the child exits the slide, you can use the principle of conservation of energy.
First, let's break down the problem and identify the different forms of energy involved.
1. Gravitational Potential Energy (PEg): When the child is at the top of the slide, all of the potential energy is in the form of gravitational potential energy. As the child slides down, this potential energy is converted into other forms.
2. Kinetic Energy (KE): As the child slides down the slide, energy is converted from potential energy to kinetic energy. At the bottom of the slide, all the potential energy is converted into kinetic energy.
3. Work Done due to Friction (Wfr): As the child slides down the slide, there is friction acting opposite to the motion. This friction does negative work, converting some of the energy into heat.
Now, the formula for conservation of energy can be written as:
PEg = KE + Wfr
The potential energy at the top of the slide is given by:
PEg = m * g * h
where m is the mass of the child, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the vertical height of the slide (3.5 m).
The kinetic energy at the bottom of the slide is given by:
KE = (1/2) * m * v²
where v is the velocity of the child at the bottom of the slide.
The work done due to friction can be calculated using:
Wfr = μk * m * g * d
where μk is the coefficient of kinetic friction (0.25), m is the mass of the child, g is the acceleration due to gravity, and d is the distance traveled along the slide.
At the bottom of the slide, the child's potential energy is zero, so the equation becomes:
0 = (1/2) * m * v² + μk * m * g * d
Now, let's rearrange the equation to solve for the velocity v:
v = √[(2 * g * h) / (1 + 2 * μk * d)]
Substituting the given values:
v = √[(2 * 9.8 * 3.5) / (1 + 2 * 0.25 * d)]
v = √[(68.6) / (1 + 0.5 * d)]
Now, you can substitute the given angle of the slide (28.0 degrees) to calculate the value of d. Since we know that the vertical height of the slide is 3.5 m, we can use trigonometry to find the distance d.
d = h / sin(θ)
d = 3.5 / sin(28.0)
d = 7.37 m (approximately)
Now, substitute this value of d back into the equation for v to find the speed at which the child exits the slide:
v = √[(68.6) / (1 + 0.5 * 7.37)]
v ≈ 4.65 m/s
Therefore, the child exits the slide with a speed of approximately 4.65 m/s.