A mass of 5 kg with initial velocity 16 m/s travels through a wind tunnel that e

Question

A mass of 5 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 4 N for a distance 2.5 m. It climbs a frictional incline of height 2.7 m inclined at an angle 17 ? , then moves along a second frictional surface of coefficient 0.18 before coming to rest. The acceleration of gravity is 9.8 m/s 2 .If the first frictional surface has a coefficient of 0.22 for a distance 0.5 m, how far does it slide along the second frictional region before coming to rest? Answer in units of m

Explanation / Answer

Given that

A mass of 5 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 4 N for a distance 2.5 m.

It climbs a frictional incline of height 2.7 m inclined at an angle 17 , then moves along a second frictional surface of coefficient 0.18 before coming to rest.

The acceleration of gravity is 9.8 m/s 2 .If the first frictional surface has a coefficient of 0.22 for a distance 0.5 m

Now the initial kinetic energy is given by

KEi =(1/2)59kg*(16m/s)2=640J

The work done by the wind tunnel (W) =4N*2.5m =10J

Therefore the total energy is givenby Etotal =650J

Now the energy lostin climbing the incline is =5*9.81*2.7m=132.435J

and the energy lost in the friction isgiven by =0.22*5*9.81*cos17*0.5=5.159J

Now the energy left in the form of mass is =650J-132.435J-5.159J =512.405J

The distance does it slide along the second frictional region before coming to rest isgiven by

=0.18*5*9.81*d =512.405J

Then d =512.405J/0.18*5*9.81 =58.036m

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