Answer:
a) [tex]W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J[/tex]
b) [tex]W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J[/tex]
c) [tex]V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s[/tex]
d) [tex]d_1 =0.183m[/tex] or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:
[tex]W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J[/tex]
Part b
For this case first we can convert the spring constant to N/m like this:
[tex]2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}[/tex]
And the work donde by the spring on this case is given by:
[tex]W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J[/tex]
Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:
[tex] W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i[/tex]
And if we solve for the initial velocity we got:
[tex]V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s[/tex]
Part d
Let d1 represent the new maximum distance, in order to find it we know that :
[tex]-1/2mV^2_i = W_g + W_{spring}[/tex]
And replacing we got:
[tex]-1/2mV^2_i =mg d_1 -1/2 k d^2_1[/tex]
And we can put the terms like this:
[tex]\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0[/tex]
If we multiply all the equation by 2 we got:
[tex] k d^2_1 -2 mg d_1 -m V^2_i =0[/tex]
Now we can replace the values and we got:
[tex]200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0[/tex]
[tex]200 d^2_1 -2.058 d_1 -6.3525=0[/tex]
And solving the quadratic equation we got that the solution for [tex]d_1 =0.183m[/tex] or 18.3 cm because the negative solution not make sense.
