Answer:
The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N
Explanation:
For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.
sin 35 = Wₓ / W
cos 35 = W_y / W
Wₓ = W sin 35
W_y = W cos 35
Wₓ = 2500 9.8 sin 35
Wₓ = 14052.6 N
let's write the equations for each axis
and
Y axis
N-W_y = 0
N = W_y
X axis
F -Wₓ = m a
F = Wₓ + m a = mg sin 35 + m a
F = m (a + g sin 35)
let's calculate
F = 2500 (5 + 9.8 sin 35)
F = 26552.6 N
The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N
The average velocity can be calculated using the formula:
v = d / t
For the 1st car, the velocity is calculated
as:
v1 = 8.60 m / 1.80 s = 4.78 m / s
While that of the 2nd car is:
v2 = 8.60 m / 1.66 s = 5.18 m / s
Now we can solve for the acceleration using the formula:
v2^2 = v1^2 + 2 a d
Rewriting in terms of a:
a = (v2^2 – v1^2) / 2 d
a = (5.18^2 – 4.78^2) / (2 * 8.6)
a = 0.23 m/s
Therefore the train has a constant acceleration of about
0.23 meters per second.
Explanation:
the rate at which something occurs or is repeated over a particular period of time or in a given sample
Answer: 20 m/s
Explanation: To solve this problem we have to consider the expression of the kinetic energy given by:
Ekinetic=(1/2)*(m*v^2)
then E=0.5*30Kg*(20 m/s)^2=15*400=6000J