If The Child Always Pulls Upward, How Much Work Does He Do?
Work Questions
If it takes a horizontal force of 300 N to push a stalled automobile along a level road at constant speed, how much work must you do to push this automobile a distance of 5.0 m?
W=FΔd
F= 300 N Δd=5 m
W=(300)(5)
W=1500 J
In an overhead lift, a champion weightlifter raises 254 kg from the floor to a height of 1.98 m. How much work does he do?
W=FΔd=Eg
Eg=mgΔh m=254 kg Δh=1.98 m
Eg=(254)(9.8)(1.98)
Eg=49528.62 J
W=49528.62 J
A child drags a 20-kg box across a lawn for 10 m and along a sidewalk for 30 m the coefficient of friction is 0.25 for the first part of the trip and 0.55 for the second. If the child always pulls horizontally, how much work does the child do on the box?
WTotal=F1Δd1+F2Δd2
F1=FN=μmg Δd1=10 m m=20 kg μ=0.25 W1=(0.25)(20)(9.8)(10) …show more content…
Ff=μmg
Ff=0.9(1200)(9.8)
Ff=10584 N
W=FΔd
W=(10584)(35.43)
W=374991.12 J
A small aircraft of mass 1200 kg is cruising at 250 km/h at an altitude of 2000 m.
What is the gravitational potential energy (relative to the ground), and what is the kinetic energy of the aircraft? m=1200 kg v=69.44 m/s Δh=2000 m
Eg=mgΔh
Eg=(1200)(9.8)(2000) Eg=23520000 J Ek=½mv2 Ek=½(1200)(69.44)2 Ek=2893148.16 J
If the pilot puts the aircraft into a dive, what will be the gravitational potential energy, what will be the kinetic energy, and what will be the speed when the aircraft reaches an altitude of 1500 m? Assume that the engine of the aircraft compensates the friction force of air, so the aircraft is effectively in free fall. m=1200 kg v=120.92 m/s Δh=1500 m
Eg=mgΔh
Eg=(1200)(9.8)(1500) Eg=17640000 J vf2=vi2+2aΔd vf2=(69.44)2+2(9.8)(500) vf2=4821.91+9800 vf=120.92 m/s Ek=½mv2 Ek=½(1200)(120.92)2 Ek=8772987.84
If it takes a horizontal force of 300 N to push a stalled automobile along a level road at constant speed, how much work must you do to push this automobile a distance of 5.0 m?
W=FΔd
F= 300 N Δd=5 m
W=(300)(5)
W=1500 J
In an overhead lift, a champion weightlifter raises 254 kg from the floor to a height of 1.98 m. How much work does he do?
W=FΔd=Eg
Eg=mgΔh m=254 kg Δh=1.98 m
Eg=(254)(9.8)(1.98)
Eg=49528.62 J
W=49528.62 J
A child drags a 20-kg box across a lawn for 10 m and along a sidewalk for 30 m the coefficient of friction is 0.25 for the first part of the trip and 0.55 for the second. If the child always pulls horizontally, how much work does the child do on the box?
WTotal=F1Δd1+F2Δd2
F1=FN=μmg Δd1=10 m m=20 kg μ=0.25 W1=(0.25)(20)(9.8)(10) …show more content…
Ff=μmg
Ff=0.9(1200)(9.8)
Ff=10584 N
W=FΔd
W=(10584)(35.43)
W=374991.12 J
A small aircraft of mass 1200 kg is cruising at 250 km/h at an altitude of 2000 m.
What is the gravitational potential energy (relative to the ground), and what is the kinetic energy of the aircraft? m=1200 kg v=69.44 m/s Δh=2000 m
Eg=mgΔh
Eg=(1200)(9.8)(2000) Eg=23520000 J Ek=½mv2 Ek=½(1200)(69.44)2 Ek=2893148.16 J
If the pilot puts the aircraft into a dive, what will be the gravitational potential energy, what will be the kinetic energy, and what will be the speed when the aircraft reaches an altitude of 1500 m? Assume that the engine of the aircraft compensates the friction force of air, so the aircraft is effectively in free fall. m=1200 kg v=120.92 m/s Δh=1500 m
Eg=mgΔh
Eg=(1200)(9.8)(1500) Eg=17640000 J vf2=vi2+2aΔd vf2=(69.44)2+2(9.8)(500) vf2=4821.91+9800 vf=120.92 m/s Ek=½mv2 Ek=½(1200)(120.92)2 Ek=8772987.84