CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A force acts on a $$30$$gm 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?


A
5.28J
loader
B
450mJ
loader
C
490mJ
loader
D
530mJ
loader

Solution

The correct option is A $$5.28$$J
$$M=20gm=0.020kg$$

$$x=3t-4t²+t³m$$

$$v=\frac { dx }{ dt } =3-8t+3t²\frac { m }{ s } $$

$$a=\frac { dv }{ dt } =-8+6t\frac { m }{ { s }^{ 2 } } $$

Force$$=m\ast a=0.120t-0.160N$$

Workdone$$=dW=Fdx=F\ast \frac { dx }{ dt } \ast dt=(0.120t-0.160)\ast (3-8t+3t²)dt=0.360t³-1.440t²+1.640t-0.480$$

$$W=0.090t⁴-0.490t³+0.82t²-0.480t=2.88J$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image