A car weighing 1.0 x 10^4N is parked on a 34 degree slope. The brakes fail and the car starts to slide down the hill. Assume no friction. Show your work, and determine a) the acceleration of the car, and b) after it has moved 28.0 m, how fast is it moving?
Why did the car go down the hill? Because it couldn't find a parking spot uphill.
Now, let's get serious (or at least try to).
To solve this problem, we need to break down the force vectors acting on the car. We have the weight of the car acting vertically downwards, which can be calculated using the formula F = mg, where m is the mass of the car and g is the acceleration due to gravity (9.8 m/s^2). In this case, the weight is given as 1.0 x 10^4 N.
Next, we need to break down the weight vector into its components. We can do this by finding the components of the weight vector that are parallel and perpendicular to the slope. The component parallel to the slope will cause acceleration, while the perpendicular component will balance out due to the lack of friction.
The component of the weight parallel to the slope can be calculated using the formula F_parallel = mg * sin(theta), where theta is the angle of the slope. So, F_parallel = (1.0 x 10^4 N) * sin(34 degrees).
Now, we can use Newton's second law, F = ma, to calculate the acceleration of the car. Since the only force acting on the car is the parallel component of the weight, we can write the equation as F_parallel = ma. Rearranging the equation, we get a = F_parallel / m.
So, the acceleration of the car is a = [(1.0 x 10^4 N) * sin(34 degrees)] / m.
To calculate the velocity of the car after it has moved 28.0 m, we can use the equations of motion. For simplicity, let's assume the car starts from rest. The equation we can use is v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (0 m/s since the car starts from rest), a is the acceleration, and s is the displacement.
Substituting the values, we get v^2 = (0 m/s)^2 + 2 * [(1.0 x 10^4 N) * sin(34 degrees) / m] * 28.0 m. Solving for v gives us the final velocity of the car.
Now, let me calculate all that boring stuff...
To determine the acceleration of the car on the slope, we can break down the weight of the car into its components.
1. First, find the force of gravity acting parallel to the slope. This can be calculated using the formula:
F_parallel = mg * sin(θ)
Where:
m = mass of the car (in kilograms)
g = acceleration due to gravity (approximately 9.8 m/s^2)
θ = angle of the slope (34 degrees)
2. The weight of the car is given as 1.0 x 10^4 N. To find the mass, use the relation:
F_weight = mg
Solving for m, we get:
m = F_weight / g
Substituting the given values, we find:
m = (1.0 x 10^4 N) / (9.8 m/s^2)
m ≈ 1020.4 kg
3. Now we can plug in the values of m and θ into the equation for F_parallel:
F_parallel = (1020.4 kg) * (9.8 m/s^2) * sin(34°)
F_parallel ≈ 5661.7 N
4. Since there is no friction, the only force acting on the car parallel to the slope is the force of gravity. Therefore, this force also determines the net force acting on the car:
F_net = F_parallel = 5661.7 N
5. Finally, we can calculate the acceleration of the car using Newton's second law:
F_net = ma
Solving for a:
a = F_net / m
Substituting the known values:
a = (5661.7 N) / (1020.4 kg)
a ≈ 5.54 m/s^2
Therefore, the acceleration of the car down the hill is approximately 5.54 m/s^2.
To determine the speed of the car after it has moved 28.0 m, we can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity (which is 0 m/s since the car started from rest)
a = acceleration (5.54 m/s^2)
s = displacement (28.0 m)
Since the initial velocity is 0 m/s, the equation simplifies to:
v^2 = 2as
Plugging in the values:
v^2 = 2 * (5.54 m/s^2) * (28.0 m)
v^2 = 310.816 m^2/s^2
Taking the square root of both sides, we get:
v ≈ 17.62 m/s
Therefore, after it has moved 28.0 m, the car is moving at approximately 17.62 m/s.
To determine the acceleration of the car on a slope, we need to consider the forces acting on it. In this case, the force of gravity is acting straight down, and we need to resolve it into components parallel and perpendicular to the slope.
1. Calculate the gravitational force parallel to the slope:
F_parallel = m * g * sin(theta)
where m is the mass of the car and theta is the angle of the slope.
Since weight = m * g, and the weight of the car is given as 1.0 x 10^4 N, we can solve for the mass:
m = weight / g
m = 1.0 x 10^4 N / 9.8 m/s^2
Substituting the calculated mass into the formula for parallel force, we have:
F_parallel = (1.0 x 10^4 N / 9.8 m/s^2) * 9.8 m/s^2 * sin(34 degrees)
2. Determine the acceleration of the car:
Since we are assuming no friction, the force parallel to the slope is the only force acting on the car. Therefore, we can relate this force to the car's mass and acceleration using Newton's second law:
F_parallel = m * a
Rearranging the equation, we get:
a = F_parallel / m
Plugging in the calculated values, we have:
a = (1.0 x 10^4 N / 9.8 m/s^2) * 9.8 m/s^2 * sin(34 degrees) / (1.0 x 10^4 N / 9.8 m/s^2)
3. Calculate the velocity of the car after it has moved 28.0 m:
We can use the kinematic equation to find the final velocity of the car after it has traveled a certain distance. The equation is:
v^2 = u^2 + 2 * a * s
Where:
v is the final velocity,
u is the initial velocity (assumed to be zero since the car is at rest),
a is the acceleration (calculated in the previous step), and
s is the displacement (28.0 m in this case).
To find the final velocity, we rearrange the equation and solve for v:
v = sqrt(2 * a * s)
Plugging in the values, we can calculate the final velocity of the car.
By following these steps, you should be able to determine both the acceleration of the car and its velocity after it has moved 28.0 m down the slope.
a. m*g = 1.0*10^4 N.
m = 1.0*10^4/g = 1.0*10^4/9.8 = 1020 kg.
Fp = 1.0*10^4*sin34 = 5592 N. = Force
parallel to the hill.
a = Fp/m = 5592/1020 = 5.48 m/s^2.
b. V^2 = Vo^2 + 2a*d = 0 + 10.96*28 = 307
V = 17.5 m/s.