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
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?
3 answers
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
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!
0 answers