A force acts on a 30g particle in such a way that the position of the particle as a function of time is given by x= 3t-4t^2 + t^3, where x is in metres and t is in seconds. The work done during the first 4 second is
Answers
Answered by
276
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
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
kvnmurty:
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Answered by
20
Answer:
5.28
Explanation:
ANSWER
The mass is 30g or 0.03kg and the position of the particle is x=3t−4t
2
+t
3
.
The velocity is given as,
dt
dx
=3−8t+3t
2
The acceleration is given as,
dt
2
d
2
x
=−8+6t
The work done is given as,
dW=(ma)dx
dW=(0.03)×(−8+6t)(3−8t+3t
2
)dt
W=(0.03)∫
0
4
(18t
3
−72t
2
+82t−24)dt
W=(0.03)(18×
4
t
4
−72×
3
t
3
+82×
2
t
2
−24t)
0
4
W=(0.03)×176
W=5.28J
Thus, the work done during the first 4s is 5.28J
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