(a) The acceleration of the system is determined as 1.58 m/s².
(b) The relative weight of P is pounds is determined as 0.14 lb.
<h3>
Acceleration of the system</h3>
The acceleration of the system is calculated as follows;
W - T = m₂a --- (1)
T = m₁a ----(2)
μmgsinθ - m₁a = m₂a
(0.3 x 3 x 9.8 x sin40) - (0.4 + 0.2)a = 3a
5.67 - 0.6a = 3a
5.67 = 3.6a
a = 5.67/3.6
a = 1.58 m/s²
<h3>
Relative Weight of P</h3>
W = ma
W = 0.4 x 1.58
W = 0.632 N = 0.14 lb
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Answer:
156.96 N
Explanation:
F=ma where m is the mass and a is acceleration
Substituting 16 Kg for m and 9.81 m/s2 for g then
F=16*9.81= 156.96 N
I believe the correct
form of the energy function is:
u (x) = (3.00 N)
x + (1.00 N / m^2) x^3
or in simpler
terms without the units:
u (x) = 3 x +
x^3
Since the
highest degree is power of 3, therefore there are two roots or solutions of the
equation.
Since we are to
find for the positions x in which the force equal to zero, u (x) = 0,
therefore:
3 x + x^3 = u
(x)
3 x + x^3 = 0
Taking out x:
x (3 + x^2) = 0
So one of the
factors is x = 0.
Finding for the
other two factors, we divide the two sides by x and giving us:
x^2 + 3 = 0
x^2 = - 3
x = sqrt (- 3)
x = - 1.732 i, 1.732
i
The other two
roots are imaginary therefore the force is only equal to zero when the position
is also zero.
Answer:
x = 0
