In compressing the spring by 0.5 m, Cho stores
1/2 (25 N/m) (0.5 m)² = 3.125 J
of potential energy. As the spring relaxes, some of this potential energy is converted into kinetic energy K in the box and the rest is lost to heat H due to friction:
K - H = 3.125 J
By the work-energy theorem, the total work W performed on the box during this process is equal to the change in the box's kinetic energy. It starts at rest when the spring is compressed, so
W = K - 0 = 1/2 (1.5 kg) v²
By Newton's second law,
• the net vertical force on the box is
∑ F (ver) = n - mg = 0
where n is the magnitude of the normal force from contact with the table, and mg is the weight of the box. It follows that
n = mg = (1.5 kg) (9.8 m/s²) = 14.7 N
• the net horizontal force is
∑ F (hor) = r - f = ma
where r is the mag. of the restoring force, f = 0.2 n is the mag. of kinetic friction, and a is the resulting acceleration of the box. Friction exerts an opposing force of
f = 0.2 (14.7 N) = 2.94 N
The work done by the spring as it relaxes is 3.125 J. The work done by friction is (2.94 N) (0.5 m) = 1.47 J. Therefore the total work done on the box is
W = 3.125 J - 1.47 J = 1.655 J
Find the speed of the box v when it returns to the equilibrium position:
1/2 (1.5 kg) v² = 1.655 J
v² = (3.31 J) / (1.5 kg)
v ≈ 1.49 m/s
I assume the rest of the table is not scarred. (If it is, the box would come to a stop well before 2.30 seconds have passed, or before the box can slide for 2.30 m.) Then the box leaves the table with speed v at an angle of 75°, so that the horizontal distance x that it covers in the air after t seconds is
x = v cos(75°) t
and its tolerable height y is
y = 2.5 m + v sin(75°) t - g/2 t²
(if y turns out negative for some t, that means it has fallen more than 2.5 m and cannot land safely)
The pile is 3.5 m away from the table, so with its launch speed the box travels this distance in time t such that
v cos(75°) t = 3.5 m
t = (3.5 m) / ((1.49 m/s) cos(75°)) ≈ 9.1 s
After such time, the box would have height
2.5 m + v sin(75°) (9.1 s) - g/2 (9.1 s)² ≈ -390.51 m
i.e. it would be nearly 400 m underground! So no, the box will not land safely.
