Your light truck weighs 3,500 kg. You drive up a gentle slope of 0.5 km. At the top of the slope, you are 200 m. assuming no friction, what force has your engine exerted to drive up this slope.
a. 7000N
b. 12000N
c. 14000N
d. 20000N
2006-11-11 00:08:47 · 3 answers · asked by babydoll53175 2 in Science & Mathematics ➔ Other - Science
Mass is 3500 kg. Distance moved is 200 meters (up).
Force is usually calculated as mass times acceleration. The SI unit used to measure force is the newton (symbol N), which is equivalent to kg*m/s^2. The acceleration here is the acceleration due to gravity or
g = 9.80665 m/s2
The force = m*g = 3500 kg * 9.81 m/s^2 = 34335 Newtons.
The energy (or work done) is the force times the height or
34335 Newtons * 200 meters = 686700 joules.
To calculate the power required, you would have to know how long the movement took in seconds.
2006-11-11 00:18:15 · answer #1 · answered by Richard 7 · 70⤊ 0⤋
The answer is C.
Work done= force*distance moved in the direction of the force(in this case, the height is in the same direction of the force)
= (3500*10)*200
(cos force=weight of truck= mass*g = 3500*10)
= 7000 000 J
Work done = force exerted to drive up*distance in the same direction (in this case, it's the slop)
7000 000 = F * 500
(cos 0.5 km=500 m)
F = 7000 000/500
= 14000 N
2006-11-12 08:12:28 · answer #2 · answered by mini_gal 3 · 0⤊ 0⤋
C
2006-11-11 12:20:50 · answer #3 · answered by Yuppy10 2 · 1⤊ 1⤋