1.

A vehicle starting from rest attainsa speed of 72 km/h after coveringa distance of 100 m. If the mass ofthe vehicle is 500 kg, the forceexerted by the engine is

Answer»

u = 0, v = 72 km/hr = 72*5/18 = 20 m/s, s = 100 m

Using,v = u + atat = 20......... (1)

Now,s = ut + 1/2 at^2100 = 0 + 1/2*(at)*t100 = 1/2*20*tt = 100/10 = 10 s

From eq(1)a*10 = 20a = 20/10 = 2 m/s^2

Force exerted by engine= mass*acceleration = 500*2 = 1000 N

v=72km/h=72*5/18=20m/su=0 s=100m v^2-u^2=2asa=v^2-u^2/2s =20^2-0^2/2*100 =400-0/200 =400/200 =2m/s^2m=500kga=2m/s^2F=ma =500*2 =1000N



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