An inclined plane of effective length 4.5m is used to raise a load of 500N through a height of 1.5m .If the list is raise by force of 200N calculate frictional force between load and the plane

1850 N

Explanation:

The formula for friction force between the load and plane is given as ;

F= μ*N

N = mg cos θ

To find θ, which is the angle the inclined plane makes with the ground at the height of 1.5 m

Sin θ = 1.5/4.5

Sin θ = 0.3333

Sin⁻{0.3333} = 19.50°

θ = 19.50°

Finding N , where m= 500 N , and g= 9.81

N = mg cos θ

N= 500 * 9.81 * cos 19.50°

N= 4624 N

Coefficient of kinetic friction is calculated as;

μ=F/W

μ = 200/500 = 0.4

The magnitude of kinetic friction is given as;

Fk= μ * N

Fk = 0.4 * 4624

Fk= 1850 N

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down