Physics, 06.03.2020 08:27 beepbeep123

A block with mass m = 4.90 kg is placed against a spring on a frictionless incline with angle θ = 45° (The block is not attached to the spring). The spring, with spring constant k = 35.0 N/cm, is compressed 16.0 cm and then released. a.) What is the elastic potential energy of the compressed spring?b.) What is the change inn the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline?c.) How far along the inline is the highest point from the release point?

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Answer from: cmaya

a) 44.8J

b) 44.8J

c) 127cm

Explanation:

a) The elastic potential energy of a compressed spring is given by the formula:

$U_e=\frac{1}{2} kx^{2}$

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:

$U_e=\frac{1}{2} (35.0N/cm)(16.0cm)^{2} = 4480Ncm$

To express the result in Joules, we have to use the fact that 1cm=0.01m. Then:

U_e=4480Ncm=4480N(0.01m)=44.8J

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:

$E_o=E_f\\\\U_{eo}+U_{go}+K_o=U_{ef}+U_{gf}+K_f$

Taking t=0 the moment in which the block is released, and t=t_f the point of its maximum height, we have that $U_{g0}=0;K_0=0;U_{ef}=0;K_f=0$ 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:

$U_{gf}=U_{e0}\\\\U_{gf}=44.8J$

So, the gravitational potential energy of the block in its highest point is 44.8J.

c) Using the gravitational potential energy formula, we have:

$U_g=mgh\\\\\implies h=\frac{U_g}{mg} \\\\h=\frac{44.8J}{(4.90kg)(9.8m/s^{2}) }=0.9m=90cm$

Using trigonometry, we can compute the distance between the release point and its highest point:

$d=\frac{h}{sin\theta}=\frac{90cm}{sin45\°}=127cm$

Answer from: Quest

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Answer from: Quest

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A block with mass m = 4.90 kg is placed against a spring on a frictionless incline with angle θ = 45...