Question

A force in the + x-direction with magnitude F(x) = 18.0N -(0.530 N/m)x is applied to a 6.00-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box. If the box is initially at rest at x=0, use the work-energy theorem to determine its speed after it has travelled 14.0 m?

Answer 1

Answer:v=8.17 m/s

Explanation:

Given

F(x)=18-0.530 x

mass of box=6 kg

distance traveled is 14 m

Work done by force

$W=\int_{0}^{14}Fdx$

$W=\int_{0}^{14}(18-0.53x)dx$

$W=\left ( 18x-\frac{0.530}{2}x^2\right )_0^{14}$

$W=252-51.94=200.06 J$

Work done by the force is equal to change in kinetic energy

$W=\frac{mv^2}{2}$

$200.06\frac{6\times v^2}{2}$

$v=8.17 m/s$