Question

A block of mass 10 kg, moving in x direction with a constant speed of 10 ms⁻¹, is subject to a retarding force F = 0.1 × J/m during its travel from x = 20 m to 30 m. Its final KE will be:(a) 450 J(b) 275 J(c) 250 J(d) 475 J

Answer 1

Using work energy theorem -

W = Kf - Ki

Kf = W + Ki

= F.dx + 1/2*m*v2

Put the values

The ans should be 475 J .

Answer 2

Answer:

The final kinetic energy is 475 J.

(d) is correct option

Explanation:

Given that,

Mass of block = 10 kg

Speed = 10 m/s

Force = 0.1 J/m

The distance travel x = 20 m to x = 30 m.

We need to calculate the final kinetic energy

Using work energy theorem

$W = \Delta K.E$

$W=K.E_{f}-\dfrac{1}{2}mv^2$

$K.E_{f}=W+\dfrac{1}{2}mv^2$

$K.E_{f}=\int_{x_{1}}^{x_{2}}{F. xdx}+\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E_{f}=\int_{20}^{30}{-0.1\times x dx}+\dfrac{1}{2}\times10\times(10)^2$

$K.E_{f}=-0.1\times(\dfrac{x^2}{2})_{20}^{30}+\dfrac{1}{2}\times10\times(10)^2$

$K.E_{f}=-0.1\times(\dfrac{30^2}{2}-\dfrac{20^2}{2})+\dfrac{1}{2}\times10\times(10)^2$

$K.E_{f}=475\ J$

Hence, The final kinetic energy is 475 J.