Answer:
The spring stretched by x = 13.7 cm
Explanation:
Given data
Mass = 3 kg
k = 120 
Angle
= 34°
From the free body diagram
Force acting on the box = mg sin
⇒ F = 3 × 9.81 × 
⇒ F = 16.45 N ------- (1)
Since box is attached with the spring so a spring force also acts on the box.
= k x
= 120
-------- (2)
The net force acting on the body is given by
Since acceleration of the box is zero so



Put the values from equation (1) & (2) we get
16.45 = 120
x = 0.137 m
x = 13.7 cm
Therefore the spring stretched by x = 13.7 cm
Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

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Explanation:
The gravitational force equation is the following:

Where:
G = Gravitational constant = 
m1 & m2 = the mass of two related objects
r = distance between the two related objects
The problem gives you everything you need to plug into the formula, except for the gravitational constant. Let me know if you need further clarification.