No net force and the photo shows
Power = Work done / time
Work done = Force * Distance
= 300 N * 1.5 m = 450 J
Power = 450 / 0.75 = 600 Watts.
Answer:
(a) the high of a hill that car can coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h is 47.6 m
(b) thermal energy was generated by friction is 1.88 x
J
(C) the average force of friction if the hill has a slope 2.5º above the horizontal is 373 N
Explanation:
given information:
m = 750 kg
initial velocity,
= 110 km/h = 110 x 1000/3600 = 30.6 m/s
initial height,
= 22 m
slope, θ = 2.5°
(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?
according to conservation-energy
EP = EK
mgh = 
gh = 
h = 
= 47.6 m
(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?
thermal energy = mgΔh
= mg (h -
)
= 750 x 9.8 x (47.6 - 22)
= 188160 Joule
= 1.88 x
J
(c) What is the average force of friction if the hill has a slope 2.5º above the horizontal?
f d = mgΔh
f = mgΔh / d,
where h = d sin θ, d = h/sinθ
therefore
f = (mgΔh) / (h/sinθ)
= 1.88 x
/(22/sin 2.5°)
= 373 N
Answer:
Ep = 3924 [J]
Explanation:
To calculate this value we must use the definition of potential energy which tells us that it is the product of mass by the acceleration of gravity by height.

where:
Ep = potential energy [J] (units of Joules)
m = mass = 40 [kg]
g = gravity acceleration = 9.81 [m/s²]
h = elevation = 10 [m]
![E_{p} =40*9.81*10\\E_{p} = 3924 [J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D40%2A9.81%2A10%5C%5CE_%7Bp%7D%20%3D%203924%20%5BJ%5D)