Kinetic energy = 1/2 m v^2 = 1/2 x1.5 x10^-3 x 0.36
Given:
Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.
We need to determine the maximum shear stress developed in the beam:
τ = F/A
Assuming the area of the beam is 100 m^2 with a length of 10 m.
τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
<span>You can use the equation
V_xf = V_xi + a_x(t)
V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0
t
Thus, solve for t and get 10seconds
and then take 5 seconds to break after 20 seconds of driving
so for
a) 10 + 20 + 5 = 35 seconds
</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
Need to find xf, knowing xi = 0
Thus, use the formula
x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
x_f = 100m
so for the first 10 seconds the truck traveled 100ms
At a speed of 20m/s
20m/s = xm/20s
20*20 = x
x = 400
thus we have 100+400 = 500m
then it slows down from 500m to x_f
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5)
x_f = 500 + 50
x_f = 550
therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s
thus
V_xavg = 16m/s</span>
