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

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

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

$v^2 = 10.8$

$v = 3.28 m/s$

Answer 2

The block's speed when it has been pulled to x=1.4m is v = 3.28 m/s

• The calculation is as follows:

According to the work energy theorem we know that total work done is equivalent to change in kinetic energy of the system.

$\frac{1}{2}mv^2 = \int\limits^ {} \, F.dx - \mu\ mgx\\\\\frac{1}{2} (2.6)(v^2) = \int\limits^ (20 -5)dx - (0.25) (2.6) ( 10) (1.4 - 0)\\\\1.3v^2 = 20 (14 - 0) - 2.5 ( 14^2 - 0) - 0.25 (2.6) ( 10) (14)\\\\v^2 = 10.8$

v = 3.28

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