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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.


​Determine the amount of energy transfer by work, in kJ, for this process and the total distance, in m, that the system travels.

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Answer:

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

Step-by-step 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=(1)/(2)mv^2\\K.E=(1)/(2)* 8* 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=(6400)/(25)=256\ m

So, the system will travel 256 m.

User Farouq Jouti
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