Initially, mg = kx. K = mg/x = 700/0.5x10^-3 = 1400000N/m. From second condition, applying work-energy theorem, potential enery- elastic potential energy = change in kinetic energy. Now change in kinetic energy is 0 since initial and final velocities are 0m/s. Therefore, potential energy = elastic potential energy. mgh = (1/2) * k* x^2. x^2 = 2(mg)h/k = 2 x 700 x 1.3/ 1400000. x = 0.036m. Hope it's clear.
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
Answer:
you could go 12 miles paying $7.80 and $1.75
So in total being $9.55
The center of mass is given with this formula:
Velocity is:
So, for the velocity of the center of mass we have:
In our case it is:
