Question

A 3.8kw elective motor powers an inclined conveyer belt. It is designed to lift heavy boxes from the warehouse floor to loading bay. Answer the following:
a)State the energy conversion performed by the motor
b) Calculate the work done by the motor in 14 s
c) If the conveyer belt takes 14 to lift its load vertically by 5.3m, calculate the maximum mass that can carried by the conveyer belt maximum mass that can be carried by the conveyer belt
d) If a small box were to fall off the conveyer belt and hit the floor, calculate the speed at which it would hit the ground if falls from a vertical height of 4.7m

Answer 1

Answer:

See the answer below

Explanation:

To solve this problem we must use the definition of power and work in physics.

a)

The function of the conveyor belt is to carry the boxes from an initial point that is at low altitude to an end point that is at high altitude. In this way the conveyor belt prints a speed to the box to be able to raise it to the required vertical distance.

Since we have a velocity at the beginning and then we place the box at a high position, where then the box remains at rest, we can say that it converts kinetic energy to potential energy.

b)

Power is defined as the relationship of work over time. Therefore we have:

$P=W/t$

where:

P = power = 3.8 [kW] = 3800 [W]

W = work [J]

t = time = 14 [s]

$W=P*t\\W=3800*14\\W= 53200[J] = 53.2[kJ]$

c)

Since the given time is equal to the given time at Point b, we can use the same work calculated.

We know that work is defined as the product of force by the distance traveled.

$W =F*d$

So, the force is equal to:

$F=W/d\\F=53200/5.3\\F=10037.73[N]$

Now we know that force is defined as the product of mass by gravitation acceleration.

$F =m*g$

where:

F = force or weight = 10037.73 [N]

g = gravity acceleration = 9.81 [m/s²]

m = mass [kg]

$m=F/g\\m = 10037.73/9.81\\m = 1023.2 [kg]$

d)

This part can be solved by means of the energy conservation theorem, where the potential energy is transformed into kinetic energy or vice versa.

$E_{pot}=m*g*h = E_{kin}=0.5*m*v^{2}$

where:

h = elevation = 4.7 [m]

v = velocity [m/s]

$m*g*h=0.5*m*v^{2}\\g*h=0.5*v^{2} \\v=\sqrt{\frac{g*h}{0.5} } \\v=\sqrt{\frac{9.81*4.7}{0.5} } \\v = 9.6 [m/s]$