At the moment \(t=0\), a time dependent force
\[
F=at
\]
where \(a\) is a constant equal to \(1\,\text{N s}^{-1}\), is applied to a body of mass \(1\,\text{kg}\) resting on a smooth horizontal plane as shown in the figure. If the direction of this force makes an angle \(45^\circ\) with the horizontal, then the velocity of the body at the moment it leaves the plane is \((g=10\,\text{m/s}^2)\):
Show Hint
For a body leaving a horizontal plane, use the condition
\[
N=0
\]
which means the upward component of applied force becomes equal to the weight.
Step 1: Understand the condition for leaving the plane.
The body leaves the horizontal plane when the normal reaction becomes zero.
The vertical component of force is
\[
F\sin45^\circ
\]
At the instant of leaving the plane,
\[
F\sin45^\circ=mg
\]
Given,
\[
F=at
\]
and
\[
a=1\,\text{N s}^{-1}
\]
So,
\[
F=t
\]
Step 2: Find the time at which the body leaves the plane.
Using
\[
F\sin45^\circ=mg,
\]
we get
\[
t\cdot \frac{1}{\sqrt{2}}=1\times 10
\]
\[
t=10\sqrt{2}\,\text{s}
\]
Step 3: Find the horizontal force responsible for motion.
The horizontal component of force is
\[
F\cos45^\circ
\]
Since
\[
F=t,
\]
the horizontal force is
\[
F_x=t\cos45^\circ
\]
\[
F_x=\frac{t}{\sqrt{2}}
\]
Step 4: Find horizontal acceleration.
Since mass is
\[
m=1\,\text{kg},
\]
the acceleration is
\[
a_x=\frac{F_x}{m}
\]
\[
a_x=\frac{t}{\sqrt{2}}
\]
Step 5: Find velocity at the instant of leaving.
The body starts from rest, so
\[
v=\int_0^{10\sqrt{2}} a_x\,dt
\]
\[
v=\int_0^{10\sqrt{2}}\frac{t}{\sqrt{2}}\,dt
\]
\[
v=\frac{1}{\sqrt{2}}\left[\frac{t^2}{2}\right]_0^{10\sqrt{2}}
\]
\[
v=\frac{1}{2\sqrt{2}}(10\sqrt{2})^2
\]
\[
v=\frac{1}{2\sqrt{2}}\times 200
\]
\[
v=\frac{100}{\sqrt{2}}
\]
\[
v=50\sqrt{2}\,\text{m/s}
\]
Step 6: Final conclusion.
Hence, the velocity of the body at the moment it leaves the plane is
\[
\boxed{50\sqrt{2}\,\text{m/s}}
\]
