18.
A force acts on a 30 g particle in such a way that
the position of the particle as a function of time
is given by x = 3t - 4t2 + t, where x is in metres
and t is in seconds. The work done during the first
4 second is :-
(1) 5.28 J
(2) 450 mJ
(4) 530 m)
(3) 490 mJ
Answers
Answer:
Explanation:
m = 20 gm = 0.020 kg
x = 3 t - 4 t² + t³ m
v = dx/dt = 3 - 8 t + 3 t² m/s
a = dv/dt = -8 + 6 t m/s²
Force = m a = 0.120 t - 0.160 N
Work done = dW = F dx = F * (dx/dt) * dt
m = 20 gm = 0.020 kg
x = 3 t - 4 t² + t³ m
v = dx/dt = 3 - 8 t + 3 t² m/s
a = dv/dt = -8 + 6 t m/s²
Force = m a = 0.120 t - 0.160 N
Work done = dW = F dx = F * (dx/dt) * dt
dW = (0.120 t - 0.160) * (3 - 8 t + 3 t²) dt
= 0.360 t³ - 1.440 t² + 1.640 t - 0.480
W = integral of dW from t = 0 to 4 sec:
= 0.090 t⁴ - 0.490 t³ + 0.82 t² - 0.480 t ] , t = 0 to 4s
= 2.88 J
Read more on Brainly.in - https://brainly.in/question/1114846#readmore= (0.120 t - 0.160) * (3 - 8 t + 3 t²) dt
= 0.360 t³ - 1.440 t² + 1.640 t - 0.480
W = integral of dW from t = 0 to 4 sec:
= 0.090 t⁴ - 0.490 t³ + 0.82 t² - 0.480 t ] , t = 0 to 4s
= 2.88 J