Q:A car accelerates uniformly from rest to 20.0 m/s in 5.6 s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is 9.0 x 10^3 N, and (b) the weight of the car is 1.4 x 10^4 N.

Ans : Read the problem note the given physical quantities with symbols from it. Given that Initial velocity u=o m/sec , Final velocity v=20m/sec , Time t = 5.6 sec , let g=10m/ .

Acceleration of the car a =(v-u)/t =(20-0)/5.6 =(20/5.6)m/

a) Mass of the car m=

m=(9.0 x )/10 ; m = 900 kg

Force on car F = = = N = 3214.28N

Average power P = ; P= = 64285.6watt or 64.2856 kwatt.

b)Mass of the car m = (1.4 x )/10 =1400kg

Force on car F = = = N = 5000N

Average power P = ; P= = watt or 100kwatt.

Q : Please define and/or explain impulse. Please say more than just product of F and t, and also the change of momentum. (Question by Tim C in Yahoo answers)

Ans : i) The effect of a large force acting on a body for a very short duration of time is called Impulse. ii) Change in momentum of a body is called Impulse iii ) The product of force applied on a body and the time for which it is applied is called Impulse. i.e Impulse = F x t (X is not vector product it is normal multiplication )

But if you try to understand the term Impulse from the equation Impulse = F x t , you will be confused. Because from this equation we come to a conclusion that, Impulse is directly proportional to time ( t ) i.e longer the period of time more will be the Impulse.Actually the Impulse will be less if the time period increases.

Smaller the time larger will be the impulse.

Ex: i)When a cricketer hits the ball hard, the time of contact (time of force applied ) of the ball is very, very short .Hence the effect will be large.More harder he hits the ball , the time of contact decreases further ,therefore the Impulse will further increase.

ii)when a person jumps on a hard surface from certain height,the moment his feet touches the floor they will be brought to rest , i.e the time of application of force is very small.Hence, the effect(Impulse) is high.But, if he jumps from the same height in to sand the foot will come to rest after some time, in this case even though the force is same but the time of application of force is increased.Hence, the effect of the force(Impulse) will be less.

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A body cover 12m in 2sec and 20m in 4sec. find what distance the body will cover in 4sec after 5th second?

(1) What modification should be done in other to increase the least count of the vernier ?

(2) what is the least count of the stop clock you are supplied with ?

Please sir answer send.