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A child’s toy consists of three cars that are pulled in tandem on small frictionless rollers as shown in below figure. The cars have masses m1 = 3.1 kg, m2 = 2.4 kg, and m3 = 1.2 kg. If they are pulled to the right with a horizontal force P = 6.5 N, find (a) the acceleration of the system, (b) the force exerted by the second car on the third car, and (c) the force exerted by the first car on the second car.
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since they lie on a frictionless track and since all 3 blocks are connected by a string the acceleration in each of them will be same
so F=Ma
6.5= 6.7×a
6.5/6.7=65/67=0.97m/s^2
Force exerted by the second car on third will occur due to string
so In F.B.D of 2
Let T1 be force exerted by car 1 on car 2
T2 be force exerted by car 3 on car 2
T1 - T2 = 2.4×0.97=2.33N
similarly T2 = 1.2xa=1.2×0.97=1.16
therefore
force exerted by second car on third car will be
T2 = 1.16 N
and that by First on second will be T1
so
2.33+1.16= 3.49 N
Answer
1)0.97m/s^2
2)1.16N
3)3.49N
so F=Ma
6.5= 6.7×a
6.5/6.7=65/67=0.97m/s^2
Force exerted by the second car on third will occur due to string
so In F.B.D of 2
Let T1 be force exerted by car 1 on car 2
T2 be force exerted by car 3 on car 2
T1 - T2 = 2.4×0.97=2.33N
similarly T2 = 1.2xa=1.2×0.97=1.16
therefore
force exerted by second car on third car will be
T2 = 1.16 N
and that by First on second will be T1
so
2.33+1.16= 3.49 N
Answer
1)0.97m/s^2
2)1.16N
3)3.49N
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