Question #3123

John and Janet were moving a heavy crate (with a mass M=250 kg) from their leveled driveway into their garage. Janet was pushing the crate horizontally with a force equal to 1/5 her weight. Her mass is 51 kg. John was pulling the crate with a 500N force at a 30 degree angle with the horizontal. The driveway and the wooden crate have a frictional forcve of 50.0N between them.
a.) total work done on the crate when it moved 45 m.
b.) It took 120 seconds to move the crate, calculate power.

Expert's answer

a) Work a=F*s, F=force, s- distance

Total work si& A=A(Janet)+A(John)-A(friction)

Hence we have

A=(51/5*9.81+500*cos(30)-50)*45=21738 Joules

b) Power P=A/t

Hence P=21738/120=181.2 Joules/s=181.2 Watt

Total work si& A=A(Janet)+A(John)-A(friction)

Hence we have

A=(51/5*9.81+500*cos(30)-50)*45=21738 Joules

b) Power P=A/t

Hence P=21738/120=181.2 Joules/s=181.2 Watt

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