Answer:
a) V = 0.82m/s
b) Vmax = 0.985 m/s
Explanation:
By conservation of energy we know that:
Eo = Ef 
Solving for V we get:
V = 0.82 m/s
To find the maximum speed we will do the same to an intermediate point where the compression is X and the distance for the work donde by frictions is given by (Xmax - X) = (0.28m - X):
Then we have to solve for V, derive and equal zero in order to find position X. After solving the derivative we get:
X = 0.1m Replacing this value into the equation for Vmax:
Vmax = 0.985m/s
Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
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Answer:
= 2630.6 N.m
Explanation:
(FR)x = ΣFx = -F4 = -407 N
(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N
(MR)B =ΣM + Σ(±Fd)
= MA + F1(d1 +d2) + F2d2 - F4d3
= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)
= 2630.64 N.m (counterclockwise)
