A spring with a force constant of 100.0N/m is used to send a box sliding across a horizontal surface. The box, with a mass of 55.0g, is pushed against the horizontal spring, compressing it 11.0cm from equilibrium. The system is then released.
(a) what is the speed of the box at the instant the spring is passing the point of being compressed 7.0cm? (it has moved 4cm away from the point of maximum compression)
(b) What is the maximum sped of the box and where does it occur?
(c) The box losses contact with the spring at maximum speed. if an average force of friction of 1.0N brings the box to a stop in 2.0s, how far will the box have travelled after leaving the spring?
(a) We can use conservation of energy to find the speed of the box at 7.0cm from the point of maximum compression. At the point of maximum compression, all the initial potential energy stored in the spring is converted to kinetic energy of the box. Therefore, we can write:
(1/2)mv^2 = (1/2)kx^2
where m is the mass of the box, v is its speed, k is the force constant of the spring, and x is the displacement of the spring from equilibrium (i.e. 11.0cm). Solving for v, we get:
v = sqrt((k/m) * x^2) = sqrt((100.0 N/m) * (0.11 m)^2 / 0.055 kg) = 2.75 m/s
Now, at 7.0cm from maximum compression, the spring is compressed by 11.0cm - 7.0cm = 4.0cm. We can again use conservation of energy to find the speed of the box at this point, since no energy is lost due to friction. Using the same equation as before with x = 0.07m, we get:
v = sqrt((100.0 N/m) * (0.07 m)^2 / 0.055 kg) = 2.31 m/s
(b) The maximum speed of the box occurs at the point where the spring is fully extended and no longer exerting any force on the box. At this point, all the initial potential energy stored in the spring is again converted to kinetic energy of the box. Therefore, we can write:
(1/2)mv^2 = (1/2)kx^2
where x is now 0.11m, the maximum compression of the spring. Solving for v, we get:
v = sqrt((k/m) * x^2) = sqrt((100.0 N/m) * (0.11 m)^2 / 0.055 kg) = 3.77 m/s
(c) After leaving the spring, the box continues to slide due to its inertia, until it is brought to a stop by friction. The average force of friction is 1.0N, so we know that the work done by friction in stopping the box is:
W = F * d = 1.0 N * d
where d is the distance travelled by the box after leaving the spring. We can use the work-energy principle to find d:
(1/2)mv_max^2 - (1/2)mv^2 = W
where v_max is the maximum speed of the box we found in part (b). Solving for d, we get:
d = (v_max^2 - v^2)/(2F) = (3.77 m/s)^2 / (2 * 1.0 N * 0.055 kg) = 6.0 m
Therefore, the box will travel 6.0m after leaving the spring before coming to a stop due to friction.