Question

What force is needed to accelerate a 680 kg object at 4.5 m/s² coefficient of

friction acting on the object is 0.65?

Answer 1

Force:

F = 7480 N

Step-by-step explanation:

Given:

m = 680 kg

a = 4.5 m/s²

μ = 0.65

g = 10 m/s²

___________

F - ?

Friction force:

F₁ = μ*m*g

Newton's Second Law:

m*a = F - F₁

m*a = F - μ*m*g

Force:

F = m*a + μ*m*g

F = m*(a + μ*g)

F = 680*( 4.5 + 0.65*10) = 680*(4.5 + 6.5) = 680*11 = 7480 N