    Question

# A cart loaded with sand having total mass 900 kg moves on a straight horizontal road under the action of a force 60 N. If cart is starting from rest and sand spills through a small hole at the bottom of cart at a rate of 0.25 kg/s, what will be the speed of cart after 10 minutes? Given: g=10 m/s2, ln(56)≈−0.2

A
50 m/s
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B
68 m/s
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C
24 m/s
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D
48 m/s
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Solution

## The correct option is D 48 m/s Initial mass of cart mi=900 kg Rate of spilling of sand =0.25 kg/s ⇒ Rate of change of mass of cart: ∴dmdt=0.25 kg/s Total mass of sand spilled after time t s is 0.25t Hence mass of cart after time t s : mf=(900−0.25t) kg From Newton's 2nd law of motion: F=mdvdt where m= mass of cart at any given time ⇒60=(900−0.25t)dvdt ⇒60∫t=600t=0dt(900−0.25t)=∫v=vv=0dv where the speed of cart is v after time 10 min or 600 s ; at time t=0,v=0 ⇒v=60∫6000dt(900−0.25t)=60−0.25[ln(900−0.25t)]6000 ∴v=−240[ln(900−0.25t)]6000 ∴v=−240×[ln(900−150)−ln(900−0)] ⇒v=−240×ln(750900)=−240×ln(56) ∴v=48 m/s  Suggest Corrections  1      Related Videos   How Do Rockets Fly in Space?
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