Question

A force pointing in the xx-direction is given by F=ax3/2F=ax3/2, where aa is a constant. The force does 2.01 kJkJ of work on an object as the object moves from xx = 0 to xxx = 15.2 mm. Find the constant aa.

Answer 1

Answer:

Explanation:

Given that

F=ax^3/2. a is a constant

The force does a work of

W=2.01KJ from x=0 to x=15.2m

We need to find a

Work is give as,

W=∫F.ds

But this is in x direction only then,

W=∫Fdx. from x=0 to x=15.2m

W=∫ax^3/2dx from x=0 to x=15.2m

W=ax^(3/2+1)/(3/2+1).

W=ax^(5/2)/5/2

W=ax^(2/5)/2.5 from x=0 to x=15.2m

Cross multiply

2.5W=ax^2.5. from x=0 to x=15.2m

2.5W= a (15.2^2.5-0)

W=2.01KJ=2010J

2.5×2010=a×900.76

Therefore,

a=5.56