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94 P

Expert-verifiedFound in: Page 292

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 ${\overrightarrow{F}}_{ext}$ 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 ${\overrightarrow{F}}_{ext}$ , (d) the work done by ${\overrightarrow{F}}_{ext}$ 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.

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

**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.**

**(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.

**(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.

**(c)**

The external force is given as:

${\overrightarrow{F}}_{ext}=\frac{dP}{dt}$

Substitute all the values in the above equation.

${\overrightarrow{F}}_{ext}=3.75N$

Therefore, the external force is 3.75 N.

**(d)**

The work done by external force is given as:

$W={\overrightarrow{F}}_{ext}\xb7v$

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

**(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.

**(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.

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