A solid sphere of mass \(2\,\text{kg}\) is at rest inside a cube as shown in the figure. The cube moves with velocity
\[
\vec v=(5t\,\hat i+2t\,\hat j)\,\text{ms}^{-1},
\]
where \(t\) is time in seconds. If the sphere is at rest with respect to the cube, find the force exerted by the sphere on the cube. (All surfaces are smooth and \(g=10\,\text{ms}^{-2}\)).
Show Hint
When an object remains at rest relative to an accelerating container, it must have the same acceleration as the container. Apply Newton's second law separately along each direction and then combine the normal reactions vectorially.
Concept:
Since the sphere remains at rest relative to the cube, it must have the same acceleration as the cube.
The sphere is in contact with the left wall and the bottom surface. Since all surfaces are smooth, only normal reactions act.
Step 1: Find the acceleration of the cube.
Given,
\[
\vec v=(5t\,\hat i+2t\,\hat j).
\]
Therefore,
\[
\vec a=\frac{d\vec v}{dt}
=
5\hat i+2\hat j.
\]
Hence,
\[
a_x=5\,\text{ms}^{-2},
\qquad
a_y=2\,\text{ms}^{-2}.
\]
Step 2: Apply Newton's second law along the \(x\)-direction.
Let \(N_1\) be the normal reaction exerted by the left wall on the sphere.
\[
N_1=ma_x.
\]
\[
N_1=2\times5.
\]
\[
N_1=10\ \text{N}.
\]
Step 3: Apply Newton's second law along the \(y\)-direction.
Let \(N_2\) be the normal reaction exerted by the floor on the sphere.
Vertical forces:
\[
N_2-mg=ma_y.
\]
\[
N_2-20=2\times2.
\]
\[
N_2=24\ \text{N}.
\]
Step 4: Find the resultant force exerted by the sphere on the cube.
By Newton's third law, the sphere exerts equal and opposite forces on the wall and floor.
Hence the resultant force on the cube is
\[
F
=
\sqrt{N_1^2+N_2^2}.
\]
\[
=
\sqrt{10^2+24^2}.
\]
\[
=
\sqrt{100+576}.
\]
\[
=
\sqrt{676}.
\]
\[
=26\ \text{N}.
\]
Therefore,
\[
\boxed{F=26\ \text{N}}
\]
\[
\boxed{\text{Answer = (C)}}
\]
