a car accelerate at 9 m/s squared. Assuming the car starts from rest how far will it travel in 10 seconds
1 answer:
You might be interested in
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²