Position Pantical of mas 2kg Depends on time (t) x=(t² - 4t)m Than Find total work Done by net Force acting on the portical in first a 2 second
Answers
Answer:
16J
Step-by-step explanation:
Given that,
Mass M=2kg
Displacement = x
Time t=2s
Now, the displacement is
x=
3
t
3
On differentiate
dt
dx
=
3
3t
2
dt
v=t
2
dt
Again differentiate
dt
dv
=2tdt
a=2tdt
Now, the force is
F=ma
F=2×2t
F=4t
Now, the work done is
dW=F⋅dx
dW=4t×t
2
dt
dW=4t
3
dt
Work done during t = 0 to 2 sec is found by integration
∫dW=
0
∫
2
4t
3
dt
W=[(2)
4
−0]
W=16J
Hence, the work done is 16 J
Answer:
16 J
Step-by-step explanation:
The relation between displacement x and time t for a body of mass 2Kg moving under the action of a force is given by x= 3t 3 ,
where x is in meter and t in second, calculate the work done by the body in first 2 seconds.
Medium
Solution
Given that,
Mass M=2kg
Displacement = x
Time t=2s
Now, the displacement is
x= 3t 3
On differentiate
dtdx = 33t dt
v=t 2 dt
Again differentiate
dtdv =2tdt
a=2tdt
Now, the force is
F=ma
F=2×2t
F=4t
Now, the work done is
dW=F⋅dx
dW=4t×t 2 dt
dW=4t 3 dt
Work done during t = 0 to 2 sec is found by integration
∫dW= 0∫24t 3 dt
W=[(2) 4 −0]
W=16J
Hence, the work done is 16 J