Question

A system with a mass of 8 kg, initially moving horizontally with a velocity of 40 m/s, experiences a constant horizontal force of 25 N opposing the direction of motion. As a result, the system comes to rest.
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​Determine the amount of energy transfer by work, in kJ, for this process and the total distance, in m, that the system travels.

Answer 1

Answer:

The kinetic energy is 6.4 kJ and the distance traveled by the system is 256 m.

Explanation:

Given the mass of the system $(m)$ is 8 kg.

And initially, it moves with a velocity $(v)$ 40 m/s.

Also, it experiences 25 N force $(f)$ which opposes its motion.

We need to find the kinetic energy and the distance traveled by the system $(d)$ before going to rest.

It will be the kinetic energy of 8 kg mass with 40 m/s velocity that is transferred to work.

$K.E=\frac{1}{2}mv^2\\K.E=\frac{1}{2}\times 8\times 40^2\\K.E= 6400\ J\\K.E=6.4\ kJ$

Since this system is opposed by 25 N force, work done by the force will be.

$W=f(d)$

And the kinetic energy transferred to work. We can equate them.

$f(d)=6400\\25(d)=6400\\d=\frac{6400}{25}=256\ m$

So, the system will travel 256 m.