consider the motion of the mass parallel to the incline
v₀ = initial velocity at the bottom of incline = 0 m/s
v = final velocity at the top of incline = 8.00 m/s
a = acceleration
d = displacement = L = length of incline = 15 m
using the equation
v² = v²₀ + 2 a d
8² = 0² + 2 a (15)
64 = 30 a
a = 64/30
a = 2.13 m/s²
F = applied force
from the force diagram, perpendicular to incline , force equation is given as
N = mg Cos30
μ = Coefficient of friction = 0.426
frictional force acting on the mass is given as
f = μ N
f = μ mg Cos30
parallel to incline , force equation is given as
F - f - mg Sin30 = ma
F - μ mg Cos30 - mg Sin30 = ma
inserting the values
F - (0.426 x 40 x 9.8) Cos30 - (40 x 9.8) Sin30 = 40 (2.13)
F = 425.82 N
Answer:
Distance, d = 778.05 m
Explanation:
Given that,
Force acting on the car, F = 981 N
Mass of the car, m = 1550 kg
Initial speed of the car, v = 25 mi/h = 11.17 m/s
We need to find the distance covered by car if the force continues to be applied to the car. Firstly, lets find the acceleration of the car:
Let d is the distance covered by car. Using second equation of motion as :
So, the car will cover a distance of 778.05 meters.