A force , F, of 500 N is applied at an angle of 37 degrees to the horizontal to

Question

A force , F, of 500 N is applied at an angle of 37 degrees to the horizontal to a mass of 100.0 kg on rough level surface and the mass is?

moved 6.0 m to the right.

A) If the mass starts from rest (vi=o) and moves at a final speed of 3 m/s after moving the 6.0 m what is the final kinetic energy of the mass?

b) What is the change in Kinetic Energy of the block?
C) From the work Kinetic enrgy theorem, how much net work is done on the mass?
D) calculate the work done by the force F=500 N if the force is applied at an angle of 37 degree to the horizontal?
E) How much work must friction be doing?

Explanation / Answer

A)final kinetic energy=0.5MV^2=0.5*100*3*3=450 J(ans)

B)change in kinetic energy=450 J (ans)

C)net work done=450J(ans)

D) net work done by force=F.d=500cos(37)*6=2400 J (ans)

E) work done by frction=2400-450 =1950 J (ans)

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