Question

A truck is moving with 5000 kg sand filled in it, there is a leakage in sand at a rate of 2 kg per minute. What should be the opposing force applied by the truck to maintain its speed at 45 km/hr (Answer Correctly with Step to Step Explanation)
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Answer 1

Given:

A truck is moving with 5000 kg sand filled in it, there is a leakage in sand at a rate of 2 kg per minute.

To find:

Opposing force necessary to maintain a speed of 45 km/hr.

Calculation:

First of all ,let's convert 45 km/hr to m/s unit :

45 km/hr = 45 × (5/18) = 12.5 m/s.

Now, we know that force is defined as rate of change of momentum :

$\rm \therefore \: f = \dfrac{dP}{dt}$

$\rm \implies \: f = \dfrac{d(mv)}{dt}$

$\rm \implies \: f = ( m \times \dfrac{dv}{dt}) +( v \times \dfrac{dm}{dt} )$

$\rm \implies \: f = \bigg \{m \times \dfrac{d(constant)}{dt} \bigg\}+( v \times \dfrac{dm}{dt} )$

$\rm \implies \: f = \bigg \{m \times 0\bigg\}+( v \times \dfrac{dm}{dt} )$

$\rm \implies \: f = v \times \dfrac{dm}{dt}$

$\rm \implies \: f = 12.5 \: m/s \times 2 \: kg/min$

$\rm \implies \: f = 12.5 \: m/s \times \dfrac{2}{60} \: kg/s$

$\rm \implies \: f = \dfrac{25}{60} \: kgm/ {s}^{2}$

$\rm \implies \: f = 0.41 \: N$

So, opposing force needed is 0.41 N.