please show work An object with mass m = 4.0 kg begins at the left side of the t
ID: 1434090 • Letter: P
Question
please show work
An object with mass m = 4.0 kg begins at the left side of the track shown at x = 0. Between x = 0 and x = 0.1 m, the object experiences a net force to the right F(x) = A(x-x_0)^2, where A is a constant and X_0 = 0.05 m. Elsewhere on the track there are no frictional forces. The object is initially at rest at x = 0 and is allowed to move under the influence of the force F(x). What is the minimum value of A such that the object flies off the track? If A = 2x10^5 N/m^2, what speed do you expect the object to hit the ground? You may assume that the effects of air resistance are negligible. If the object hits the ground with speed 3.0 m/s, how much energy is lost to air resistance?Explanation / Answer
a) Lets first calculate work done by force,
dW = F.dx = A(x - x0)^2 dx
W = A (x - x0)^3 / 3
from x = 0 to x
W = A [(x - x0)^3 - (0 -x0)^3 ] = A [ (x - x0)^3 + x0^3 ]
if work done is more than mgh then block will reach at the top:
A [ (x - x0)^3 + x0^3 ] = mgh
A [ (0.1 - 0.05)^3 + (0.05)^3 ] = 4 x 9.8 x 0.25
A = 39200 N/m^2
b) Work done = 2 x 10^5(2 x 0.05^3) = 50 J
speed at top: 50 = mgh + mv^2 /2
50 = (4 x 9.8 x0.25) + (4 v^2 / 2)
2v^2 = 40.2
v = 4.48 m/s
Now using energy conservation to find speed at the ground.
(4 x 4.48^2 /2 ) + ( 4 x 9.8 x 0.75) = (4 v^2 /2 ) + 0
v = 5.90 m/s
c)energy lost to air = change in KE
= 5 ( 5.90^2 /2 - 3^2 /2 )
= 64.43 J
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