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InterviewSolution
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A car weighing 1000 kg and travelling at `30 m//s` stops at a distance of 50 m decelerating uniformly. What is the force exerted on it by the breaks ? What is the work done by the brakes? |
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Answer» In order to calculate the force, we have to find out the acceleration (or rather retardation) first. Now, Initial speed, u = 30 m/s Final speed, v = 0 (The car stops) Acceleration, a = ? (To be calculated) And, Distance, s = 50 m Now, we know that : `v^(2) = u^(2) + 2as` So, `(0)^(2) = (30)^(2) + 2 xx a xx 50` 100 a = - 900 `a = -(900)/(100)` Thus, Acceleration, `a = - 9 m//s^(2)` The force exerted by the brakes can now be calculated by using the formula: `F = m xx a` Here, Mass, `m = 1000 kg " "` (Given) And, Acceleration, `a = -9 m//s^(2)" "` (Calculated above) So, Force, `F = 1000 xx (-9)` F = - 9000 N Thus, the force exerted by the brakes on the car is of 9000 newtons. The negative sign shows that it is a retarding force. The work done by the brakes can be calculated by using the relation : `W = F xx s` Here, Force, F = 9000 N Distance, s = 50 m So, Work done, `W = 9000 xx 50 J` = 450000 J `= 4.5 xx 10^(5) J` Thus, the work done by the brakes is `4.5 xx 10^(5)` joules. |
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