Question

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.90 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, what is its speed after it has traveled 15.0 m ?

Answer 1

Answer:

6.875 m/s

Explanation:

The force is variable which is given by

F(x) = 18 - 0.53 x

mass of the box, m = 8.9 kg

initially it is at rest at x = 0

Let the velocity is v after travelling a distance of 15 m.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

$\int \overrightarrow{F}.d\overrightarrow{x}=\Delta K.E$

$\int_{0}^{15} \left ( 18-0.53 x \right )dx=\frac{1}{2}\times m \left ( v^{2}-u^{2} \right )$

$\left ( 18x-0.265x^{2} \right )_{0}^{15}=\frac{1}{2}\times 8.9\times \left ( v^{2}-0^{2} \right )$

18 x 15 - 0.265 x 15 x 15 = 4.45 x v²

270 - 59.625 = 4.45 v²

v² = 47.275

v = 6.875 m/s

Thus, the final velocity of the box is 6.875 m/s.