question 14 in the image In the following diagram, M = 5 kg, m = 3 kg and the kineticfrictional coefficient is 0.25 between the surface and the big M. Thesmall m is 3 meters above the ground.14. find the work done by the friction as the big M slidesthrough a distance of 4 meters.15. find the speed of the small m when it touches the ground.Mdмк6m4m

Name: question 14 in the image In the following diagram, M = 5 kg, m = 3 kg
Uploaded: 2024-03-20T06:57:25.000Z
Channel: Bartleby
Description: question 14 in the image In the following diagram, M = 5 kg, m = 3 kg and the kineticfrictional coefficient is 0.25 between the surface and the big M. Thesmall m is 3 meters above the ground.14. find the work done by the friction as the big M slidest

given The mass placed on the surface is M=5kg The hanging mass is m=3kg coefficient of kinetic friction is μ=0.25 the height from the ground of mass m is h=4m* *Considering the image is correct.14) So the friction force between mass M and the surface will be f=μN=μMg=0.25×5kg×9.8m/s2=12.25N The distance traveled by the mass M is d=4m So the work done by the frictional force will be ∆W=-fd=-12.25N×4m=-49J15) Let at the ground the velocity of the mass m is v Then its total energy at the ground is Ef=12mv2 And the total energy of mass m at the initial position is Ei=mgh So from the conservation of energy ∆W=Ef-Ei Or -49J=12mv2-mgh the velocity of the mass m at the ground is v=2mmgh-49J=23kg3kg×9.8m/s2×4m-49J=6.76m/s