A car can decelerate at -4.80 m/s^2 without skidding when coming to rest on a level road. What would its deceleration be if the road were inclined at 13 degrees uphill? Assume the same static friction force.
Supposively the answer is -7.00 m/s^2 but I keep on getting -2.47 m/s^2 because I interperted it up the hill and decelerating
nevertheless when I consider it as the car going down hill and decelerating I'm getting -6.88 m/s^2
can you please show me how to get -7.00 m/s^2
Thanks
The deceleration rate is the ratio of applied force (backwards along the direction of motion) to the mass.
The mass remains the same. Therefore the acceleration is higher by the ratio
(Fstatic + M g sin 13)/Fstatic
You also know that the decelerating force applied by the tires is
Fstatic = M*4.80,
(from Newton's law), and that it stays the same
Therefore
(deceleration-uphill)/(deceleration-level)
= 1 + M g sin 13/M*4.80
= 1 + g sin 13/4.80 = 1.4597
(deceleration-uphill) = 1.4597*(-4.80)
= -7.007 m/s^2
If I had used 9.80 instead of 9.81 m/s^2 for g, I would have gotten -7.004 as an answer
Well, it seems like you're having a rocky road with this question. Let's try to straighten things out.
When the car is on a level road, the deceleration is given as -4.80 m/s^2. Now, when the road is inclined uphill at 13 degrees, we need to determine the deceleration in this new scenario.
First, let's analyze the forces acting on the car. We have the force of gravity pulling the car downhill and the static friction force acting uphill to prevent slipping.
Now, since the car is decelerating and coming to rest, we know that the force of static friction must be equal to the force pulling the car downhill, which is the component of gravity acting parallel to the incline.
Now, we can use some trigonometry to determine this force. The component of gravity parallel to the incline is given by:
Force = mass * gravity * sin(theta)
where theta is the angle of the incline (13 degrees in this case).
Since the force of static friction is equal to the force pulling the car downhill, we can equate the two forces:
Force of static friction = mass * acceleration
Now, we know the acceleration is negative because the car is decelerating, so we have:
mass * acceleration = mass * gravity * sin(theta)
Mass cancels out on both sides, giving us:
acceleration = gravity * sin(theta)
Plugging in the values, we get:
acceleration = 9.8 m/s^2 * sin(13 degrees)
And if you calculate this, you'll find:
acceleration ≈ -2.038 m/s^2
Oh boy, it looks like my calculation doesn't match your answer of -7.00 m/s^2. To be honest, I can't see how we can magically get -7 without any other information or missing variables. It's possible there might be something else at play or an additional factor that needs to be considered.
Nevertheless, humor me for a second. Let's imagine if the car had rocket boosters hidden underneath it, then maybe, just maybe, it could magically achieve -7.00 m/s^2 deceleration uphill. But unfortunately, we're talking about real-world physics here, and unlikely scenarios aren't going to get us to the answer.
On a serious note, I would recommend taking a second look at the question and double-checking the given information and any possible missing pieces. If you have access to additional resources or instructions, that might also help clarify the intended solution.
I hope this helps, even if it's not the 7 you were looking for!
To find the deceleration of the car when the road is inclined at 13 degrees uphill, we need to consider the forces acting on the car.
When the car is on a level road, the maximum static friction force provided by the road is equal to the mass of the car multiplied by the acceleration due to gravity (mg), where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the car can decelerate at -4.80 m/s^2 without skidding on a level road, the magnitude of the maximum static friction force is equal to the product of the mass of the car and -4.80 m/s^2. Let's call this force FR (for resistance).
Now, when the road is inclined at 13 degrees uphill, we need to consider the components of the weight (mg) and the normal force acting on the car.
The weight (mg) can be split into two components: one parallel to the inclined road and one perpendicular to the inclined road.
The component of the weight parallel to the inclined road is given by mg * sin(13 degrees), and the component perpendicular to the inclined road is given by mg * cos(13 degrees).
On the inclined road, the normal force (FN) is equal in magnitude and opposite in direction to the perpendicular component of the weight (mg * cos(13 degrees)).
Since the car can still decelerate without skidding, the maximum static friction force (FR) is equal to the magnitude of the normal force (FN).
Therefore, we can set up the following equation:
mg * sin(13 degrees) = FR = FN = mg * cos(13 degrees)
Dividing both sides of the equation by mg, we get:
sin(13 degrees) = cos(13 degrees)
Now, we can solve for the deceleration (a) on the inclined road:
a = FR / m = FN / m = g * cos(13 degrees) = 9.8 m/s^2 * cos(13 degrees) ≈ 9.8 m/s^2 * 0.978 ≈ 9.60 m/s^2
Therefore, the deceleration of the car when the road is inclined at 13 degrees uphill is approximately -9.60 m/s^2. Since the problem states that it should be -7.00 m/s^2, there may be an error in the given answer.
To understand how to calculate the deceleration of a car on an inclined road, we need to break down the forces acting on the car.
When the car is on a level road, the only force acting on it is the friction force. The friction force opposes the car's motion and causes it to decelerate until it comes to rest. In this case, the deceleration is given as -4.80 m/s^2.
Now, let's consider the car on an inclined road. When the road is inclined, the force of gravity comes into play. The force of gravity acts down the inclined road, causing the car to slide or accelerate if no other force opposes it. However, in this scenario, we assume that the same static friction force is acting on the car, opposing its motion as it decelerates.
To calculate the deceleration on the inclined road, we need to analyze the forces acting on the car in the direction perpendicular to the road's surface.
1. Resolve the force of gravity: The force of gravity can be resolved into two components - one parallel to the road's surface and one perpendicular to it. The component parallel to the road is balanced by the friction force, so it does not contribute to the acceleration.
2. Determine the perpendicular component of gravity: The perpendicular component is given by the formula F_perpendicular = mg * sin(theta), where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of the incline (13 degrees in this case).
3. Calculate the net force: The net force acting on the car is the difference between the perpendicular component of gravity (downhill) and the opposing static friction force (uphill). This net force will cause the car to decelerate.
4. Use Newton's second law: Apply Newton's second law of motion, which states that the net force is equal to the mass of the object multiplied by its acceleration. In this case, the net force is given by F_net = m * a, where m is the mass of the car and a is the deceleration.
Putting it all together:
F_perpendicular = mg * sin(theta)
Net force = F_perpendicular - Friction force
m * a = Net force
In this case, since we are looking for the deceleration, we can reorganize the equation:
a = (F_perpendicular - Friction force) / m
Substituting the given values:
m = 1 (since the mass is not specified)
theta = 13 degrees
Friction force = -4.80 m/s^2 (given on level road)
a = [(1 * 9.8 * sin(13)) - (-4.8)] / 1
a = [1.34 + 4.8] / 1
a = 6.14 / 1
a = -6.14 m/s^2 (rounded to 2 decimal places)
Therefore, the correct deceleration of the car on the inclined road is approximately -6.14 m/s^2. It seems that the answer you were given (-7.00 m/s^2) may be either rounded or calculated slightly differently.