Respuesta :
Answer:
a) 44.8J
b) 44.8J
c) 127cm
Explanation:
a) The elastic potential energy of a compressed spring is given by the formula:
[tex]U_e=\frac{1}{2} kx^{2}[/tex]
Where U_e is the elastic potential energy, k is the spring constant and x is the distance the spring is compressed. In this case, we have:
[tex]U_e=\frac{1}{2} (35.0N/cm)(16.0cm)^{2} = 4480Ncm[/tex]
To express the result in Joules, we have to use the fact that 1cm=0.01m. Then:
[tex]U_e=4480Ncm=4480N(0.01m)=44.8J[/tex]
In words, the elastic potential energy of the compressed spring is 44.8J.
b) Using the law of conservation of mechanical energy, we have that:
[tex]E_o=E_f\\\\U_{eo}+U_{go}+K_o=U_{ef}+U_{gf}+K_f[/tex]
Taking t=0 the moment in which the block is released, and t=t_f the point of its maximum height, we have that [tex]U_{g0}=0;K_0=0;U_{ef}=0;K_f=0[/tex] because in t=0 the block has no speed and is in tis lowest point; and in t=t_f the block has stopped and isn't in contact with the spring. So, our equation is reduced to:
[tex]U_{gf}=U_{e0}\\\\U_{gf}=44.8J[/tex]
So, the gravitational potential energy of the block in its highest point is 44.8J.
c) Using the gravitational potential energy formula, we have:
[tex]U_g=mgh\\\\\implies h=\frac{U_g}{mg} \\\\h=\frac{44.8J}{(4.90kg)(9.8m/s^{2}) }=0.9m=90cm[/tex]
Using trigonometry, we can compute the distance between the release point and its highest point:
[tex]d=\frac{h}{sin\theta}=\frac{90cm}{sin45\°}=127cm[/tex]
