a 1200kg car coasts from rest down a driveway that is inclined 20degrees. to the horizontal and is 15m long. how fast is the car going at the end of the delivery if....
(a) friction is negligible
(book answer is 10m/s)
(b) a friction force of 3000n opposes the motion?
(book answer is 5.1 m/s)
dont know where to begin to solve these 2questions. I tried differant formulas, but nothing I tried came out to the answers in the book.
help
Many if not most problems in physics are not solved by plugging directly into a single equation. It is better to prceed in logical steps using formulas you are familiar with.
In this case, the place to start is with conservation of energy. The vertical drop of the runway is
H = 15 sin 20 = 5.13
The Potential energy loss MgH equals the Kinetic Energy gain, so
MgH = (1/2) M V^2
V = sqrt(2 g H) = 100.3 m/s
(I used 9.81 m/s^2 for g)
Note that you did not need to use the mass for this one.
To do the second case, subtract friction work done
3000N*15 m = 45,000 J from the potential energy decrease, and solve for V again.
M g H - 45,000 = (1/2) M V^2
You will need to use the mass this time.
see this very related problem:
http://www.jiskha.com/display.cgi?id=1266960070
Two long, parallel wires separated by 30 cm each carry currents of 4.0 A in a horizontal direction.
B)Find if they are in opposite directions.
To solve these questions, we can break down the problem into two parts:
1. Determine the acceleration of the car down the incline.
2. Use the acceleration to calculate the final velocity of the car.
Let's start with part (a) where the friction is negligible:
1. Determine the acceleration of the car down the incline:
To find the acceleration, we need to resolve the force of gravity along the incline. The force of gravity can be broken down into two components: one parallel to the incline (mg*sin(20°)) and the other perpendicular to the incline (mg*cos(20°)).
The only force parallel to the incline is the component of gravity. So, we can write the equation:
m * a = m * g * sin(20°)
Where:
m = mass of the car = 1200 kg
a = acceleration down the incline
g = acceleration due to gravity = 9.8 m/s² (approximate value)
Simplifying the equation gives us:
a = g * sin(20°)
2. Calculate the final velocity of the car:
Now that we have the acceleration, we can use one of the kinematic equations to determine the final velocity of the car. The equation we can use is:
v^2 = u^2 + 2*a*s
Where:
v = final velocity of the car (what we want to find)
u = initial velocity of the car (which is 0 m/s because the car starts from rest)
a = acceleration down the incline (found in step 1)
s = distance traveled down the incline = 15 m
Plugging in the values, the equation becomes:
v^2 = 0^2 + 2 * (g * sin(20°)) * 15
v^2 = 2 * 9.8 * sin(20°) * 15 (approximating g as 9.8 m/s²)
Taking the square root and solving, we get:
v ≈ 10 m/s (rounded to one decimal place)
So, the final velocity of the car when the friction is negligible is approximately 10 m/s, which matches the book answer.
Now let's move on to part (b) where a friction force of 3000 N opposes the motion:
1. Determine the acceleration of the car down the incline:
This time, we need to include the frictional force opposing the motion. The equation becomes:
m * a = m * g * sin(20°) - F_friction
Where:
F_friction = frictional force opposing motion = 3000 N
So, the equation becomes:
m * a = m * g * sin(20°) - 3000
2. Calculate the final velocity of the car:
Following the same steps as part (a), we can use the kinematic equation:
v^2 = 0^2 + 2 * (g * sin(20°) - F_friction / m) * s
Plugging in the values, the equation becomes:
v^2 = 2 * 9.8 * sin(20°) - 3000 / 1200 * 15
Solving for v gives us:
v ≈ 5.1 m/s (rounded to one decimal place)
So, the final velocity of the car when a friction force of 3000 N opposes the motion is approximately 5.1 m/s, which matches the book answer.
I hope this explanation helps you understand how to approach these types of problems!