A 1.6kg box that is sliding on a frictionless surface with a speed of 11m/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?
a. It compresses until
(1/2)MVo^2 = (1/2)kX^2
k is the spring constant.
Solve for X
b. X is proportional to Vo. Half the Vo results in half the deflection.
To find the distance the spring will be compressed in stopping the box, we can use the principle of conservation of mechanical energy. The initial kinetic energy of the box is converted into potential energy stored in the spring when it comes to rest.
a. The equation for the potential energy stored in a spring is given by:
PE = (1/2)kx^2
where PE is the potential energy stored, k is the spring constant, and x is the distance the spring is compressed.
The initial kinetic energy of the box is given by:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the box, and v is the initial velocity.
Since the initial kinetic energy is equal to the potential energy stored, we have:
(1/2)mv^2 = (1/2)kx^2
Plugging in the given values, we have:
(1/2)(1.6kg)(11m/s)^2 = (1/2)(2000 N/m)(x^2)
Simplifying the equation, we get:
96.8 = 1000x^2
Dividing both sides by 1000, we have:
x^2 = 0.0968
Taking the square root of both sides, we have:
x ≈ 0.311 meters
Therefore, the spring will be compressed by approximately 0.311 meters in stopping the box.
b. To find the distance the spring will be compressed when the box's speed is reduced to half of its initial speed, we can use the concept of conservation of mechanical energy again.
We'll use the same equation: (1/2)mv^2 = (1/2)kx^2
This time, the initial kinetic energy is given by:
KE = (1/2)mv^2
where v is the initial velocity of the box.
The final kinetic energy when the box's speed is reduced to half of its initial speed is given by:
KE' = (1/2)m(v/2)^2
Plugging in the given values, we have:
(1/2)mv^2 = (1/2)kx^2
(1/2)mv^2 = (1/2)kx'^2
Dividing both sides of the equation, we get:
v^2 = x'^2
Since the box's speed is reduced to half, we have:
(v/2)^2 = x'^2
Squaring both sides of the equation, we have:
v^2/4 = x'^2
Plugging in the initial kinetic energy equation, we have:
(v^2/4) = (1/2)kx'^2
Simplifying the equation, we get:
v^2/2 = kx'^2
Plugging in the given values, we have:
(11m/s)^2/2 = (2000 N/m)x'^2
Simplifying the equation, we get:
121/2 = 2000x'^2
Dividing both sides by 2000, we have:
x'^2 = 0.0605
Taking the square root of both sides, we have:
x' ≈ 0.246 meters
Therefore, the spring will be compressed by approximately 0.246 meters when the box's speed is reduced to half of its initial speed.