A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is dropped onto the spring.

(a) How far does the object compress the spring?

m

(b) Repeat part (a), but this time assume a constant air-resistance force of 0.750 N acts on the object during its motion.

m

(c) How far does the object compress the spring if the same experiment is performed on the Moon, where g = 1.63 m/s2 and air resistance is neglected?

m

(A) l = 0.39 m

(B) l = 0.38 m

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

potential energy of the object = mg (1.05 + l)

potential energy of the spring = 0.5 x k x l^{2} (k = spring constant)

therefore we now have

mg (1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) = 0.5 x 300 x l^{2}

15.68 (1.05 + l) = 150 x l^{2}

16.5 + 15.68l = 150l^{2}

l = 0.39 m

(B) with constant air resistance the equation applied in part A above becomes

initial P. E of the object - air resistance = final P. E of the spring

mg (1.05 + l) - 0.750 (1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) - 0.750 (1.05 + l) = 0.5 x 300 x l^{2}

(16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}

16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

15.71 + 14.93l = 150^{2}

l = 0.38 m

(C) where g = 1.63 m/s^{2} and neglecting air resistance

the equation mg (1.05 + l) = 0.5 x k x l^{2} now becomes

1.6 x 1.63 x (1.05 + l) = 0.5 x 300 x l^{2}

2.608 (1.05 + l) = 0.5 x 300 x l^{2}

2.74 + 2.608l = 150 x l^{2}

l = 0.14 m