Question

A 6.0-kg box moving at 3.0 m/s on a horizontal, friction-less surface runs into a light spring of force constant 75 N/cm. Use the work—energy theorem to find the maximum compression of the spring.

Answer 1

To find the maximum compression of the spring, we can use the work-energy theorem. The maximum compression of the spring is 0.24 m.

Step-by-step explanation:

To find the maximum compression of the spring, we can use the work-energy theorem. The work done by the box on the spring will be equal to the change in kinetic energy of the box. The work is given by the formula: work = (1/2)kx^2, where k is the force constant of the spring and x is the compression of the spring.

Since the box is initially moving and comes to rest when it compresses the spring, the change in kinetic energy is equal to the initial kinetic energy of the box. The initial kinetic energy is given by the formula: KE = (1/2)mv^2, where m is the mass of the box and v is its initial velocity.

Setting the work done equal to the initial kinetic energy and solving for x, we get: x = sqrt((2KE)/k).

Plugging in the values given in the question, we have: x = sqrt((2(6.0 kg)(3.0 m/s)^2) / (75 N/cm)). Calculating this expression gives the maximum compression of the spring as 0.24 m.

Answer 2

The maximum compression in the spring is 0.085 m or 8.5 cm.

Step-by-step explanation:

Given:

Mass of box (m) = 6.0 kg

Spring constant of spring (k) = 75 N/cm = 75 × 100 N/m = 7500 N/m

Initial velocity of the box (u) = 3.0 m/s

Final velocity at maximum compression (v) = 0 m/s

Let the maximum compression be 'x'.

Surface is frictionless. So, only spring force is the force acting on the box spring system.

Now, as per work-energy theorem, the work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

Here, the work done by the spring force is given as:

`W_(net)=-(1)/(2)kx^2`

The negative sign implies that force and displacement are in opposite direction.

The change in kinetic energy is given as:

`\Delta K=(1)/(2)m(v^2-u^2)`

Now, from work-energy theorem:

`W_(net)=\Delta K\\\\-(1)/(2)kx^2=(1)/(2)m(v^2-u^2)\\\\x^2=(m(v^2-u^2))/(-k)\\\\x=\sqrt{(m(v^2-u^2))/(-k)}`

Now, plug in the given values and solve for 'x'. This gives,

`x=\sqrt{(6(0-3^2))/(-7500)}\\\\x=\sqrt{(-54)/(-7500)}\\\\x=0.085\ m =8.5\ cm\ \ [1\ m=100\ cm]`

Therefore, the maximum compression in the spring is 0.085 m or 8.5 cm.