# POM question e ( only need part E completing) my answers are attached

A ROBOT UNLOADS A FINISHED COMPONENT FROM A MACHINE AND PLACES IT ON A GRAVITY CONVEYOR. THE CONVEYOR ACCELERATES FROM REST UNTIL IT REACHES THE BOTTOM OF THE CONVEYOR 4.5 SECONDS LATER. THE COMPONENT IS TRAVELLING AT A VELOCITY OF 1.35 M/S WHEN IT REACHES THE BOTTOM OF THE CONVEYOR WHERE , IT IS THEN STOPPED BY HITTING A BUFFER. THE COMPONENT SITS STATIONARY AWAITING PACKAGING.

A) DETERMINE THE LINEAR ACCELERATION OF THE COMPONENT AS IT TRAVELS DOWN THE GRAVITY CONVEYOR (IGNORING FRICTIONAL FORCES.

B) THE CONVEYOR SECTION LENGTH IS LATER MODIFIED AND EXTENDED BY A FURTHER 1.5M. DETERMINE THE NEW VELOCITY OF THE COMPONENTS WHEN THEY REACH THE BUFFER.

C) EXPLAIN THE CAUSE OF THE COMPONENTS ACCELERATING FROM REST DOWN THE CONVEYOR.

D) DETERMINE THE FORCE EXERTED BY THE COMPONENT, OF MASS 4.5KG, DOWNWARD ON THE CONVEYOR AS IT SITS AT THE BUFFER.

E) EXPLAIN WHY THE COMPONENT DOES NOT ACCELERATE DOWN THROUGH THE CONVEYOR WHILST RESTING AT THE BUFFER.

A)

FOR THE FIRST PART, USE EQUATION V_{F} = V_{0} + AT, WHERE V_{0} = 0, T = 4.5 S AND V_{F }= 1.35 M/S

A = 1.35/4.5 = 0.3 M/S^{2}

B)

KNOWING ACCELERATION AND NEW DISTANCE, YOU CAN DETERMINE THE NEW FINAL SPEED USING EQUATION:

(VF)^{2 }= V_{0}^{2} + 2A*D

THE PROBLEM STATES THAT IT IS EXTENDED BY AN ADDITIONAL 1.5 M, SO YOU’LL HAVE TO FIRST DETERMINE THE INITIAL LENGTH BEFORE THE EXTENSION:

D = VF^{2}/2A = (1.35)^2/(2*.3) = 3.0375 M

NOW YOU ADD 1.5 TO THAT DISTANCE ABOVE AND USE THE FORMULA TO DETERMINE V_{F}:

V_{F }= ROOT(2*.3*4.5375) = 1.65M/S

C) GRAVITY

D)

WE KNOW THAT

VELOCITY (V) = ACCELERATION (A) X TIME (T)

1.35 = A X 4.5 ———à A=0.3 M/S^{2}

FORCE (F) = MASS (M) *ACCELERATION (A)

= 4.5 KG *0.3 M/S^{2 }=1.35

**E)**

**THE FORCE EXERTED WILL BE EQUAL TO MASS TIMES THE GRAVITY. THAT IS 4.5*9.81= 44.145 N.**

**THE COMPONENT DOES NOT ACCELERATE DOWN THE CONVEYOR BELT BECAUSE WHEN ENTIRE ASSEMBLY IS CONSIDERED AS A SYSTEM, IT DOES NOT AFFECT OUR SOLUTION AND AS IT DOES NOT AFFECT THE SOLUTION. THE ACCELERATION OF ALL PARTS IS SAME AND HENCE, COMPONENT MAINTAINS ITS INERTIA UNDER THE EFFECT OF GRAVITY **