Will mark the brainliest
When a car of mass 1000kg is travelling along a level road at a steady speed of 20ms −1
its engine is working at 18kW.
Resultant force = 900N Braking Force = 3100N Gravity = 9.81
By how much does this change if the car is travelling down the same hill at 20ms −1 ?

Respuesta :

Answer:

OK draw a diagram - you have the force from (b) acting sown the slope and a component of the weight. Use F=ma to get the deceleration and then use SUVAT. Post your working if this doesn't work.

friction = 900N

braking = 3100N

total = 4000N

w= mxg = 1000*9.81= 9810N

total = 9810 + 4000 = 13810N

force/mass = 13810/1000 = 13.81ms^-2

then using v^2 = u^2 + 2as, i get s as 14 but it is incorrect

Explanation: