Answer:
36.8 N
Explanation
tension in string when both bodies move vertically is given as
T = (m1 × m2 × g) / ( m1 +m2)
T = (5 × 3 × 9.81)/(5+3)
T = 147.15/8
T = 18.39375
SINCE TENSIONS ARE ON BOTH SIDES OF PULLEY AND ON PULLEY BOTH ARE DIRECTED DOWNWARD
HENCE FORCE ON PULLEY = 2 × T
= 2×18.39375
= 36.7875
= 36.8 N
Answer:
a = 0.5195 m/s²
θ = 9.997º ≈ 10º
Explanation:
We apply Newton's 2nd Law as follows:
∑ Fx = m*ax
∑ Fy = m*ay
Then we have
∑ Fx = F₁x + F₂x = m*ax ⇒ 600*Cos 40º + 600*Cos (-20º) = 2000*ax
⇒ ax = 0.5117 m/s²
∑ Fy = F₁y + F₂y = m*ay ⇒ 600*Sin 40º + 600*Sin (-20º) = 2000*ay
⇒ ay = 0.0902 m/s²
the magnitude of the acceleration of the barge is
a = √(ax² + ay²) = √((0.5117 m/s²)² + (0.0902 m/s²))= 0.5196 m/s²
and the direction is
θ = Arctan (ay / ax) = Arctan (0.0902 / 0.5117) = 9.997º ≈ 10º
The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:

where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.
Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.
When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:

