Can someone help me understand this physics FRQ? It's my hardest subject
A lab cart starts at rest in a long hallway. A force acts horizontally on the cart for a distance of 5 meters. This force has a magnitude of 0 Newtons to start and increases uniformly to 100 Newtons at the 5 meter point in the hall. From the 5 meter point, the force remains constant and horizontal for the next 15 meters. Finally, the horizontal force decreases until it is 0 Newtons after the cart has traveled a total distance of 30 meters.
a. Create a graph of Force vs. Distance for the entire 30 meter trip.
b. The velocity of the cart is measured as 10 m/s when the cart is at a position of 5 meters from the starting point. Calculate the mass of the cart.
c. Calculate the velocity of the cart at the end of the 30 meter trip.
d. If an average frictional force of -50 Newtons is applied to the cart after it reaches the 30 meter point, what is the stopping distance for the cart?
well for starters
first 5 meters
F = k x
at 5 meters
100 = 5 k
k = 20
so
F = 20 x
a(x)= F/m = 20 x/m
v(x) = (20/m) integral x dx
v = (20/m) x^2/2
v = 10 x^2 / m
but at x = 5, v = 10
so
10 = 10 (25)/m
m = 25 kilograms
at 5 meters v = 10 m/s
next 10 meters force = 100 Newtons
a = F/m = 100 /25 = 4 m/s^2
v = 10 + a t = 10 + 4 t
x = 5 + 10 t + (4/2) t^2
at x = 15
15 = 5 + 10 t + 2 t^2
2 t^2 + 10 t - 10 = 0
t^2 + 5 t - 5 = 0
t = .854 or -5.85 from https://www.mathsisfun.com/quadratic-equation-solver.html use 0.854
v = 10 + 4 (0.854) = 13.4
now for the last part F = 10(30 - x) etc
thank you!
Sure, I can help you understand this physics FRQ. Let's go step by step:
a. To create a graph of Force vs. Distance for the entire 30 meter trip, you need to understand how force changes with respect to distance in each segment.
In the first segment, the force starts at 0 Newtons and increases uniformly to 100 Newtons at the 5 meter point. So, for the first 5 meters, the force increases linearly from 0 to 100 Newtons.
In the second segment, the force remains constant and horizontal for the next 15 meters. So, for these 15 meters, the force stays at a constant value of 100 Newtons.
Finally, in the third segment, the force decreases until it becomes 0 Newtons after the cart has traveled a total distance of 30 meters. So, for the last 10 meters, the force decreases linearly from 100 to 0 Newtons.
Now, you can plot the graph with force on the y-axis and distance on the x-axis. Connect the data points to get a complete graph of force vs. distance for the entire 30 meter trip.
b. To calculate the mass of the cart, you need to use the given information about velocity and force.
The force acting on an object is given by the equation F = ma, where F is the force, m is the mass, and a is the acceleration.
In this case, you have the velocity (v = 10 m/s) and force (F = 100 Newtons) when the cart is at a position of 5 meters from the starting point.
Since the velocity is constant, the acceleration is zero and hence the net force acting on the cart is zero at this point. Therefore, you can set up the equation 0 = 100 Newtons - F(friction), where F(friction) is the frictional force acting opposite to the motion.
Given that the average frictional force is -50 Newtons, you can solve the equation to find F(friction) = -50 Newtons.
Now, you can use F = ma and solve for mass. Plugging in the known values, you have 100 Newtons - (-50 Newtons) = ma.
Solving for mass (m), you get m = (100 Newtons - (-50 Newtons)) / a.
Since the acceleration (a) is zero, the mass of the cart is undefined or infinite. This means that either the given information is inconsistent or there is missing information to determine the mass.
c. To calculate the velocity of the cart at the end of the 30 meter trip, you need to consider the forces acting on the cart and apply the principles of Newton's second law.
In the first segment, the force is increasing uniformly and positive, so it accelerates the cart in the positive direction.
In the second segment, the force is constant and positive, so the cart continues to accelerate in the positive direction at a constant rate.
In the third segment, the force decreases and eventually becomes zero, so the cart continues moving in the positive direction at a constant velocity until it reaches the end of the trip.
Since the cart is not acted on by any external forces, the final velocity at the end of the trip is the same as its velocity during the last segment. Therefore, you can use the given information that the velocity is constant at 10 m/s for the last 10 meters to find the velocity at the end of the 30 meter trip.
d. To determine the stopping distance for the cart when an average frictional force of -50 Newtons is applied after it reaches the 30 meter point, you need to consider the forces acting on the cart and apply Newton's second law.
The force of friction opposes the motion of the cart, so it acts in the negative direction.
When the force of friction is equal to the force pushing the cart, the cart stops moving. So, you can set up the equation F(friction) = F(push) to find the distance where the cart stops.
Given that the average frictional force is -50 Newtons and the pushing force is 0 Newtons after the cart reaches the 30 meter point, you have -50 Newtons = 0 Newtons.
Solving for the stopping distance, you find that the cart stops immediately after reaching the 30 meter point.
I hope this explanation helps you understand the physics FRQ. If you have any further questions, feel free to ask!