Question

A BLOCK SLIDES DOWN AN INCLINED PLANE OF SLOPE ANGLE THETA WITH A CONSTANT VELOCITY. IT IS THEN PROJECTED UP THE PLANE WITH AN INITIAL VELOCITY U. DISTANCE UPTO WHICH IT WILL RISE BEFORE COMING TO REST IS ?

Answer 1

Answer on Question #40894 – Physics – Mechanics

A BLOCK SLIDES DOWN AN INCLINED PLANE OF SLOPE ANGLE THETA WITH A CONSTANT VELOCITY. IT IS THEN PROJECTED UP THE PLANE WITH AN INITIAL VELOCITY U. DISTANCE UPTO WHICH IT WILL RISE BEFORE COMING TO REST IS?

Solution:

$\mathrm{F_{fr}}$ – friction force;

$\theta$ – angle of the inclined plane;

$\mathrm{U}$ – initial velocity;

$t$ – time after the block stopped;

$a$ – deceleration of the block;

Newton’s second law for the block when it’s moving down along the X-axis before it was projected ($V = \text{const} \Rightarrow \text{acceleration} = 0$):

$$x: F_{fr} - mg_x = 0$$ $$F_{fr} = mg_x$$

From the right triangle ABC:

$$\sin \theta = \frac{mg_x}{mg} \Rightarrow mg_x = mg \sin \theta$$

Newton’s second law for the block when it’s moving up along the X-axis after it was projected. Force that acts of the block:

$$x: F_{net} = F_{fr} + mg_x$$

(1) and (2) in (3):

$$F_{net} = mg_x + mg_x = 2mg_x = 2mg \sin \theta$$

Loss of KE by block = work done by friction force + PE (final speed of the block = 0):

$$\frac{mU^2}{2} = F_{net} \cdot d + mgd \cdot \sin \theta$$

(4) in (5):

$$\frac{mU^2}{2} = 2mgd \sin \theta + mgd \cdot \sin \theta$$ $$\frac {U ^ {2}}{2} = 3 \mathrm {g d} \sin \theta$$ $$d = \frac {U ^ {2}}{6 g \sin \theta}$$

Answer: distance upto which block will rise before coming to rest is equal to $\frac{\mathrm{U}^2}{6\mathrm{g}\sin\theta}$ .