Unit: 3. Work , Energy and Power
Weightage : 10-15%
A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ignoring air resistance,
Correct Answer: \( \boxed{\mathrm{A}} \)
Ignoring air resistance, all objects near Earth’s surface accelerate downward at the same rate:
\( a=g \)
Therefore, choice (B) is incorrect because mass does not affect the acceleration due to gravity.
The first rock is released \(0.5\,s\) earlier and therefore always has a greater downward velocity than the second rock while both are in the air.
Since both rocks experience the same acceleration, their velocity difference remains constant. The first rock continues to move farther downward during each time interval, causing the separation between the rocks to increase.
Choice (C) is incorrect because the second rock is released exactly \(0.5\,s\) later and therefore reaches the ground approximately \(0.5\,s\) later as well.
Choice (D) is incorrect because kinetic energy depends on mass:
\( K=\frac{1}{2}mv^2 \)
Although both rocks strike the ground with the same speed, the more massive rock has twice the kinetic energy.
Therefore, the correct answer is (A).
From the top of a high cliff, a ball is thrown horizontally with initial speed \(v_0\). Which of the following graphs best represents the ball’s kinetic energy \(K\) as a function of time \(t\)?
Correct Answer: \( \boxed{\mathrm{D}} \)
The ball is thrown horizontally with an initial speed \(v_0\), so it already has kinetic energy at \(t=0\):
\( K_0=\frac{1}{2}mv_0^2 \)
As the ball falls, gravity causes its vertical velocity to increase according to
\( v_y=gt. \)
The horizontal velocity remains constant:
\( v_x=v_0. \)
Therefore, the total speed is
\( v=\sqrt{v_x^2+v_y^2} =\sqrt{v_0^2+g^2t^2}. \)
The kinetic energy becomes
\( K=\frac{1}{2}m(v_0^2+g^2t^2). \)
Since \(K\) contains a \(t^2\) term, it increases quadratically (parabolically) with time. Because the ball starts with nonzero kinetic energy, the graph begins above zero and curves upward.
Thus, the graph that best represents the kinetic energy as a function of time is (D).
An object of mass \(m\) is initially at rest and free to move without friction in any direction in the \(xy\)-plane. A constant net force of magnitude \(F\) directed in the \(+x\) direction acts on the object for \(1\,\text{s}\). Immediately thereafter, a constant net force of the same magnitude \(F\) directed in the \(+y\) direction acts on the object for \(1\,\text{s}\). After this, no forces act on the object.
Which of the following graphs best represents the kinetic energy \(K\) of the object as a function of time?
.
Correct Answer: \( \boxed{\mathrm{B}} \)
During the first second, the object experiences a constant force in the \(+x\) direction, producing a constant acceleration
$ a=\frac{F}{m}. $
Since the object starts from rest, its speed increases linearly with time:
$ v=at. $
Therefore, the kinetic energy increases as
$ K=\frac12 mv^2 =\frac12 m(at)^2, $
which is a parabola.
From \(t=1\) to \(t=2\,\text{s}\), the force acts only in the \(+y\) direction. The \(x\)-component of velocity remains constant, while the \(y\)-component increases linearly. The total speed is
$ v=\sqrt{v_x^2+v_y^2}, $
so the kinetic energy continues to increase with a curved (quadratic) shape.
After \(t=2\,\text{s}\), no forces act on the object. Its velocity remains constant, so its kinetic energy also remains constant.
The graph that matches this behavior is (B).
