Question

A box of mass m = 4 kg and spring exist on a frictionless table of height h = 0.2 meters. The spring is attached to a solid wall and is compressed a distance x = 0.06 meters by the box. When released, the box is projected off of the table, landing a distance d = 1.2 meters from the edge of the table. Calculate the following. (b) The box's speed at the instant it leaves the table.(c) The energy initially stored in the spring.(d) The spring constant.

Answer 1

Given data

*The given mass of the box is m = 4 kg

*The given frictionless table of height is h = 0.2 m

*The spring compressed at a distance is x = 0.06 m

*The table landing a distance is d = 1.2 m

(b)

The formula for the time taken by the box is given by the equation of motion as

`\begin{gathered} h=(1)/(2)gt^2 \\ t=\sqrt[]{(2h)/(g)} \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} t=\sqrt[]{(2*0.2)/(9.8)} \\ =0.202\text{ s} \end{gathered}`

The formula for the box's speed at the instant it leaves the table is given as

`v=(d)/(t)`

Substitute the known values in the above expression as

`\begin{gathered} v=(1.2)/(0.202) \\ =5.94\text{ m/s} \end{gathered}`

Hence, the box's speed at the instant it leaves the table is v = 5.94 m/s

(d)

The formula for the spring constant is given by the conservation of energy as

`\begin{gathered} U_s=U_k \\ (1)/(2)kx^2=(1)/(2)mv^2 \\ k=(mv^2)/(x^2) \end{gathered}`

*Here U_s is the spring energy

*Here U_k is the kinetic energy

Substitute the known values in the above expression as

`\begin{gathered} k=((4)(5.94)^2)/((0.06)^2) \\ =3.92*10^4\text{ N/m} \end{gathered}`

Hence, the spring constant is k = 3.92 × 10^4 N/m

(c)

The formula for the energy initially stored in the spring is given as

`U_s=(1)/(2)kx^2`

Substitute the known values in the above expression as

`\begin{gathered} U_s=(1)/(2)(3.92*10^4)(0.06)^2 \\ =70.56\text{ J} \end{gathered}`

Hence, the energy initially stored in the spring is U_s = 70.56 J