A 1.5-kg box that is sliding on a frictionless surface with a speed of 12 m/s approaches a horizontal spring. The spring has a spring constant of 2000 N/m. (a) How far will the spring be compressed in stopping the box? (b) How far will the spring be compressed when the box's speed is reduced to half of its initial speed?
Hi,
part a)
Given frictionless surface.
Mass of box, m = 1.5 kg
velocity of the box, v = 12 ms^-1
Spring constant, k = 2000 N/m
From principle of conservation of energy:
when the box is stopped by the spring, all the kinetic energy in to box will be converted to elastic potential energy inside the spring. Hence we can say that:
Kinetic energy of box = elastic potential energy of spring
where,
x = compression of the spring
substituting known values into equation (1)
part b)
when the velocity of the box is reduced to half its speed in part a)
v= 6 ms^-1
using the same calculations as in part a)
x = 0.164316767 meters = 16.4316 centimeters