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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=(9000.25t) kg

From Newton's 2nd law of motion:
F=mdvdt
where m= mass of cart at any given time
60=(9000.25t)dvdt
60t=600t=0dt(9000.25t)=v=vv=0dv
where the speed of cart is v after time 10 min or 600 s ; at time t=0,v=0
v=606000dt(9000.25t)=600.25[ln(9000.25t)]6000
v=240[ln(9000.25t)]6000
v=240×[ln(900150)ln(9000)]
v=240×ln(750900)=240×ln(56)
v=48 m/s

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