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Grade: 11
        
a force F=5t is applied on a block of mass 10 kg at rest on a rough horizontal surface [co effiecient of friction=.2]what is its speed at t=20 seconds?
2 years ago

Answers : (1)

Arun
23752 Points
							
Dear Student
 
 
acceleartion = (F-uN)/m =(5t – 0.2 * 100)/10 = (5t – 20)/2 m/s^2
a = dv/dt =(5t – 20)/2 m/s^2
dv = [(5t – 20)/2]dt
integerating both sides,
v = 5t^2/4 – 10t
at t=20s 
V = 500 - 200 = 300 m/sec
 
Regards
Arun (askIITians forum expert)
2 years ago
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