It has to be the last one because whenever lights are turned on it decreases because all lights are on at the same time. It's good to just have one light on. It doesn't use as much electricity.
We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
</span>F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2<span>μ Fn/m</span> = Δd. This is equal to
- Angle (θ) = 60°
- Force (F) = 20 N
- Distance (s) = 200 m
- Therefore, work done
- = Fs Cos θ
- = (20 × 200 × Cos 60°) J
- = (20 × 200 × 1/2) J
- = (20 × 100) J
- = 2000 J
<u>Answer</u><u>:</u>
<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
