Question

A force F=(15 +0.50x) acts on a particle in the
X-direction, where F is in newton and x in metre.
Find the work done by this force during a displace-
ment from x = 0 to x = 2.0 m. ()​

Answer 1

Answer:

the and is may be 31Nm. if ans is wrong thn plz comment.

Answer 2

Answer:

31 J

Explanation:

We are given that force acting on a particle in the x- direction is given by

$F=15+0.50 x$

Where F ( in N) and x(in meters)

We have to find the work done by the force during a displacement from x=0 to x=2 m

We know that work done =f dx

$W=\int_{0}^{2}(15+0.5x) dx$

$W=[15x+0.5\frac{x^2}{2}]^2_0$

$W=15(2)+\frac{0.5}{2}(2)^2-15(0)-\frac{0.5}{2}(0)^2$

$\int_{a}^{b}fdx=F(b)-F(a)$

$W=30+1=31 J$

Hence, the work done by the force=31 J