Question

A 2.6 kg block is attached to a horizontal rope that exerts a variable force Fx=(20−5x)N, where x is in m. The coefficient of kinetic friction between the block and the floor is 0.25. Initially the block is at rest at x = 0 m.
What is the block's speed when it has been pulled to x=1.4m?

Answer 1

Answer:

Speed of the block after it is pulled a distance x = 1.4 m is v = 3.28 m/s

Explanation:

As per work energy theorem we know that total work done is equal to change in kinetic energy of the system

so here we have

$\frac{1}{2}mv^2 = \int F.dx - \mu mg x$

so we will have

$\frac{1}{2}(2.6)v^2 = \int (20 - 5x)dx - (0.25)(2.6)(10)(1.4 - 0)$

$1.3v^2 = 20(1.4 - 0) - 2.5(1.4^2 - 0) - 0.25(2.6)(10)(1.4)$

$v^2 = 10.8$

$v = 3.28 m/s$