energyhomework-and-exercisesnewtonian-mechanics
Question: Which of the following graphs correctly describes the variation of kinetic energy with time of a block when it slides down a smooth inclined plane from rest?
The answer is C but I do not understand why. How do we know if the velocity increases as it slides down the inclined plane?
The block's $x$-velocity right before it hits the floor is not $v_b$, but $v_b \cos \alpha + v_I$. The $\cos \alpha$ is because it's not moving horizontally relative to the wedge, but at an angle $\alpha$; the $+ v_I$ term is because it's moving relative to the wedge, and momentum is only conserved in a frame that is at rest with respect to the table. Similarly, its $y$-velocity is $v_b \sin \alpha$. So the correct equations are (taking into account the relative motions): $$ \frac{1}{2} m (v_b^2 + v_I^2 + 2 v_b v_I \cos \alpha) + \frac{1}{2} M v_I^2 = mgh $$ $$ m (v_b \cos \alpha + v_I) + M v_I = 0 $$ Solving these two equations yields the expression you got from applying Newton's Second Law.
You are forgetting the normal force between the block and the wedge (inclined plane). This force has a horizontal component, pushing the wedge to the left as the block slides down to the right.
The 2 forces acting on the wedge due to the block are the normal reaction $N$ and the friction force $F=\mu |N|$ (see diagram). Both have horizontal components : $N\sin\theta$ and $F\cos\theta$ respectively. The net horizontal force acting to the left on the wedge is $N\sin\theta-F\cos\theta=(\tan\theta -\mu)N\cos\theta$.
The condition for the block to start sliding down the incline is $\tan\theta \gt \mu$. So if the block slides to the right then the horizontal force on the wedge is always +ve to the left.
Note that in this situation the normal force is not $N=mg\cos\theta$. The wedge accelerates to the left away from the plane of contact with the block, so the normal force is reduced from this value.
Best Answer
The conserved mechanical energy is:
$E=KE+V=KE+mgh$
where $h$ is the height of the body. As the body slides down due to the acceleration of gravity, the distance travelled would be $d=\tfrac{1}{2}a{{t}^{2}}$, where $a$ is the acceleration caused by the addition of the weight of the body and the vertical reaction force. If $L$ is the length of the inclined plane and $\theta $ the inclination angle, then:
$h=\left( L-d \right)\sin \theta =L\sin \theta -\tfrac{1}{2}a{{t}^{2}}\sin \theta $
Plugging this to the mechanical energy formula and rearranging the terms we get:
$KE=A+B{{t}^{2}}$
where $A=E-mgL\sin \theta $ and $B=\tfrac{1}{2}mga\sin \theta $. Since it is only the energy difference that is of physical importance, we can set the mechanical energy level such that $A=0$. So:
$KE=B{{t}^{2}}$
which is the equation for a convex parabola that crosses the origin.