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5.9 Aleration 4 kg moving iniay with a constant sothsubfect to a constant force of 8.0 Nt time to be x 0, and predict its posiA barth is ut the force is applied to be tdirectesDeedtrajectory of th(a) the stone n(b) the stone(c) the stone f5.10 A body of mass 0.40the instant the torce is applied to hethe30 s. Take theo be 0, and predict its position at t, the poste;

Answer»

Mass of the body,m= 0.40 kg

Initial speed of the body,u= 10 m/s due northForce acting on the body,F= –8.0 NAcceleration produced in the body, a = F / m = -8.0 / 0.40 = -20 ms-2

(i) Att= –5 sAcceleration,a‘ = 0 andu= 10 m/ss =ut+ (1/2) a’ t2= 10 × (–5) = –50 m

(ii) Att= 25 sAcceleration,a” = –20 m/s2andu= 10 m/ss’ = ut’ + (1/2) a” t2= 10 ×25 + (1/2)×(-20)×(25)2= 250 – 6250 = -6000 m

(iii) Att= 100 sFor 0 ≤ t ≤ 30 sa = -20 ms-2u = 10 m/ss1= ut + (1/2)a”t2= 10×30 + (1/2)×(-20)×(30)2= 300 – 9000 = -8700 mFor 30 < t ≤ 100 s

As per the first equation of motion, fort= 30 s, final velocity is given as:v=u + at= 10 + (–20) × 30 = –590 m/sVelocity of the body after 30 s = –590 m/sFor motion between 30 s to 100 s, i.e., in 70 s:s2= vt + (1/2) a” t2= -590×70 = -41300 m∴ Total distance, s” = s1+ s2= -8700 -41300 = -50000 m = -50 km.


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