Question

Sand falls on to a conveyor belt at a constant rate of 2kg/sec. If the belt is moving at 0.1m/sec. Then the extra force required to maintain speed of the belt is

Answer 1

Answer:

• The extra Force (F) required to maintain the speed of belt is 0.2 N.

Given:

1. Sand falls at a rate of ( M / t ) = 2 Kg / sec.
2. Velocity of the belt is (v) = 0.1 sec

Explanation:

$\rule{300}{1.5}$

From the Newton's Second law of motion,

⇒ F = M a

Where,

• F Denotes the Force.
• M Denotes Mass.
• a Denotes Acceleration.

Now,

⇒ F = M a

Simplifying,

⇒ F = M × v / t     ∵ [ a = v / t ]

Here, velocity is constant and mass keeps on varying, so,

⇒ F = v × M / t

Substituting the values,

⇒ F = 0.1 m/s × 2 Kg / sec.

∵ [ M / t = 2 Kg / sec ]

⇒ F = 0.1 × 2

⇒ F = 0.2

⇒ F = 0.2 N

∴ The extra Force (F) required to maintain the speed of belt is 0.2 N.

$\rule{300}{1.5}$

Answer 2

Answer:

Given:

Sand falls on conveyor belt at a rate of 2 kg/sec.

The belt is moving at 0.1 m/s

To find:

Extra force needed to maintain the speed of belt

Concept:

We know that force is the rate of change of momentum. In other words , we can say that time derivative of momentum gives force. The derivative is complex because mass also changes with time.

$P = m \times v$

$= > force = \dfrac{dP}{dt}$

$= > force = \dfrac{d(m \times v)}{dt}$

$= > force = v \dfrac{dm}{dt} + m \dfrac{dv}{dt}$

Since velocity change doesn't take place , so dv/dt = 0

$= > force = v \dfrac{dm}{dt} + (m \times 0)$

$= > force = v \dfrac{dm}{dt}$

Calculation:

$force = v \dfrac{dm}{dt}$

$= > force = 0.1 \times 2$

$= > force = 0.2 \: N$

So final answer :

$\boxed{ \red{ \huge{ \bold{ force = 0.2 \: N }}}}$