91Ó°ÊÓ

Here we have, the mass of block is\(m = 1.5\,{\rm{kg}}\).

The incline slope is\(\theta = {30^ \circ }\).

The distance between point \(A{\rm{ and }}B\) is \(d = 6.0\,{\rm{m}}\)

The speed of the block at point \(B\) is \({v_B} = 700\,{\rm{m/s}}\)

The coefficient of the kinetic friction between the block and the incline is \({\mu _k} = 0.50\)

Potential energy is a form of stored energy that is dependent on the arrangement of system components.

So, work-energy theorem is given by:

\({K_1} + {U_{grav,1}} + {U_{el,1}} + {W_{other}} = {K_2} + {U_{grav,2}} + {U_{el,2}}\) ...(1)

Here,

\({K_1},\,{K_2}\)is initial and final kinetic energy respectively.

\({U_{grav,1}},\,{U_{grav,2}}\)is initial and final gravitational potential energy respectively.

\({U_{el,1}},\,{U_{el,2}}\)is initial and final elastic potential energy respectively.

\({W_{other}}\)is work done by other forces.

Where kinetic energy is given by:

\(K = \frac{1}{2}m{v^2}\) ...(2)

Where \(m\,\,{\rm{and}}\,\,v\)is mass and velocity of objective respectively.

The gravitational potential energy is given by:

\({U_{grav}} = mgy\) ...(3)

Where \(m\,\,{\rm{and}}\,\,y\) is mass and distance respectively.

We also know that the friction force is given by:

\({f_k} = {\mu _k} \cdot n\) ...(5)

Where \({\mu _k}\) is spring constant

Since the friction force is always opposite to motion’s direction, so the work done by friction is given by:

\({W_f} = - {f_k} \cdot d = - {\mu _k} \cdot n \cdot d\) ...(6)

Where \({\mu _k}\) is spring constant and \({f_k}\) is friction force.

Now, by applying Newton’s second law to the block along the direction perpendicular to the incline, so we get:

\(\begin{aligned}{}\sum F = n - mg\cos \theta \\\sum F = 0\\n = mg\cos \theta \\n = 1.50\,{\rm{kg}} \times 9.8\,{\rm{m/}}{{\rm{s}}^2} \times \cos 30\\n = 12.73\,{\rm{N}}\end{aligned}\)

Now put value of \({\mu _k},\,n\,,\,d\)in equation (6)

\(\begin{aligned}{}{W_f} &= - \left( {0.50} \right) \cdot \left( {12.73\,{\rm{N}}} \right) \cdot \left( {6.0\,{\rm{m}}} \right)\\ &= - 38.19\,{\rm{J}}\end{aligned}\)

Free body diagram from given information is,

Let\(y = 0\)at point A.

Now for point A

\(\begin{aligned}{l}{y_A} = 0,\\{v_A} = 0\end{aligned}\)

For point B,

\(\begin{aligned}{}{y_B} &= d\sin \theta \\ = 6\sin 30,\\{v_B} &= 7.0\,{\rm{m/s}}\end{aligned}\)

Since the block starts from rest and from the zero gravitational potential energy level, then we get:

\({K_1} = 0,\,\,\,\,\,\,\,\,{U_{grav,1}} = 0\)

Also, since at point B, the block is no more in contact with the spring.

Then we get:

\({U_{el,2}} = 0\)

Now put value of \({v_B}\,{\rm{and }}m\)in equation (2)

\(\begin{aligned}{}{K_2} &= \frac{1}{2}\left( {1.50\,{\rm{kg}}} \right){\left( {7.0\,{\rm{m/s}}} \right)^2}\\ &= 36.75\,{\rm{J}}\end{aligned}\)

Now put value of \({y_B}\,{\rm{and }}m\) in equation (3)

\(\begin{aligned}{}{U_{grav,2}} &= \left( {1.50\,{\rm{kg}}} \right) \cdot \left( {9.8\,{\rm{m/}}{{\rm{s}}^2}} \right) \cdot \left( {3.0\,{\rm{m}}} \right)\\ &= 44.1\,{\rm{J}}\end{aligned}\)

Now, put these values in equation (1),

\(\begin{aligned}{}0 + 0 + {U_{el,1}} - 38.19\,{\rm{J}} &= 36.15\,{\rm{J}} + 44.1\,{\rm{J}} + 0\\{U_{el,1}} &= 119.04\,{\rm{J}}\end{aligned}\)

Hence, the potential energy that was initially stored in the spring is \(119.04\,{\rm{J}}\).