Question #85456

An object of mass 20 kg is pulled up with a force of 150 N on an inclined plane for a
distance of 3.0 m. The plane makes an angle of 30º with the horizontal and the
coefficient of kinetic friction between the object and the plane is 0.3. Draw the free
body diagram and calculate the work done by each force. (Take g = 10 ms−2
)

Expert's answer

Answer on Question #85456, Physics / Other

An object of mass m=20 kgm = 20 \, \mathrm{kg} is pulled up with a force of F=150 NF = 150 \, \mathrm{N} on an inclined plane for a distance of s=3.0 ms = 3.0 \, \mathrm{m} . The plane makes an angle of θ=30∘\theta = 30{}^\circ with the horizontal and the coefficient of kinetic friction between the object and the plane is μ=0.3\mu = 0.3 . Draw the free body diagram and calculate the work done by each force.

Solution:

The free body diagram is as follows



The work done by definition


W=Fscos⁡αW = F s \cos \alpha


We get


WF=Fscos⁡0∘=Fs=150×3.0=450JW _ {F} = F s \cos 0 {}^ {\circ} = F s = 1 5 0 \times 3. 0 = 4 5 0 \mathrm {J}Wmg=mgscos⁡(90∘+θ)=−mgssin⁡θ=−20×9.8×3.0×12=−294JW _ {m g} = m g s \cos (9 0 {}^ {\circ} + \theta) = - m g s \sin \theta = - 2 0 \times 9. 8 \times 3. 0 \times \frac {1}{2} = - 2 9 4 \mathrm {J}WN=Nscos⁡90∘=0JW _ {N} = N s \cos 9 0 {}^ {\circ} = 0 \mathrm {J}WFf=Ffscos⁡180∘=−μmgscos⁡θ=−0.3×20×9.8×3.0×32=−153JW _ {F _ {f}} = F _ {f} s \cos 1 8 0 {}^ {\circ} = - \mu m g s \cos \theta = - 0. 3 \times 2 0 \times 9. 8 \times 3. 0 \times \frac {\sqrt {3}}{2} = - 1 5 3 \mathrm {J}

Answers:

WF=150J,W _ {F} = 1 5 0 \mathrm {J},Wmg=−294J,W _ {m g} = - 2 9 4 \mathrm {J},WN=0,W _ {N} = 0,WFf=−153JW _ {F _ {f}} = - 1 5 3 \mathrm {J}


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