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Expert-verified Found in: Page 292 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # Sand from a stationary hopper falls onto a moving conveyor belt at the rate of 5.00 kg/s as shown in Figure P9.94. The conveyor belt is supported by frictionless rollers and moves at a constant speed of v = 0.750 m/s under the action of a constant horizontal external force ${\stackrel{\to }{F}}_{ext}$ supplied by the motor that drives the belt. Find (a) the sand’s rate of change of momentum in the horizontal direction, (b) the force of friction exerted by the belt on the sand, (c) the external force ${\stackrel{\to }{F}}_{ext}$ , (d) the work done by ${\stackrel{\to }{F}}_{ext}$ in 1 s, and (e) the kinetic energy acquired by the falling sand each second due to the change in its horizontal motion. (f) Why are the answers to parts (d) and (e) different?

(a) The rate of change of momentum of sand in horizontal direction is 3.75 N.

(b) The force of friction exerted by the belt on sand is 3.75 N.

(c) The external force is 3.75 N.

(d) The work done by external force is 2.812 J.

(e) The kinetic energy acquired by sand due to change in its motion is 1.41 J.

(f) The answers in part (d) are different due to friction work and work of external force.

See the step by step solution

## Step 1: Identification of given data

The rate of sand from hopper is $\frac{dm}{dt}=5kg/s$

Constant speed of conveyor belt is $v=0.750m/s$

The duration for work done by external force is t = 1s

## Step 2: Understanding the concept

The impulse momentum theorem states that impulse of an object due to force on object for small duration is equal to the change in momentum of object.

## Step 3: Determination of rate of change of momentum of sand in horizontal direction

(a)

The rate of change of momentum of sand in horizontal directionis given as:

$\frac{dP}{dt}=v\left(\frac{dm}{dt}\right)$

Substitute all the values in the above equation.

$\frac{dP}{dt}=\left(0.750m/s\right)\left(5kg/s\right)\phantom{\rule{0ex}{0ex}}\frac{dP}{dt}=3.75N$

Therefore, the rate of change of momentum of sand in horizontal direction is 3.75 N.

## Step 4: Determination of force of friction exerted by the belt on sand

(b)

The force of friction exerted by the belt on sand is given as:

$f=\frac{dP}{dt}$

Substitute all the values in the above equation.

$f=3.75N$

Therefore, the force of friction exerted by the belt on sand is 3.75 N.

## Step 5: Determination of external force

(c)

The external force is given as:

${\stackrel{\to }{F}}_{ext}=\frac{dP}{dt}$

Substitute all the values in the above equation.

${\stackrel{\to }{F}}_{ext}=3.75N$

Therefore, the external force is 3.75 N.

## Step 6: Determination of work done by external force

(d)

The work done by external force is given as:

$W={\stackrel{\to }{F}}_{ext}·v$

Substitute all the values in the above equation.

$W=\left(3.75N\right)\left(0.750m/s\right)\phantom{\rule{0ex}{0ex}}W=2.812J$

Therefore, the work done by external force is 2.812 J

## Step 7: Determination of kinetic energy acquired by sand due to change in its motion

(e)

The kinetic energy acquired by sand due to change in its motion is given as:

$\Delta K=\frac{1}{2}\left(\frac{dm}{dt}\right){v}^{2}$

Substitute all the values in the above equation.

$\Delta K=\frac{1}{2}\left(5kg/s\right){\left(0.750m/s\right)}^{2}\phantom{\rule{0ex}{0ex}}\Delta K=1.41J$

Therefore, the kinetic energy acquired by sand due to change in its motion is 1.41 J.

## Step 8: Reason for difference in answers of part (d) and (c)

(f)

The part (d) is giving the work done by the external force and kinetic energy acquired by sand is equal to the work done by external force against friction. ### Want to see more solutions like these? 