Question

A 10kg crate sits at rest on a rough flat surface. Astudent decides to pull the crate by attaching a rope at a 37 degree angle. Although the student pulls the rope with a force of 600 newtons, the coefficient of kinetic friction is large and has the force the student applies remains constant, how much time after he begins pulling the crate will it take before the crate has traveled a distance of 1.0 meter?

Answer 1

Step-by-step explanation:

Normal force of the surface on the box will be

N = mg - Fsinθ

Ν = 10(9.8) - 600sin37

N = -263

As normal force cannot be less than zero, the applied force lifts the crate off the surface.

Now it's just a matter of finding the acceleration

In the horizontal direction, the acceleration is

a = F/m

a = (600cos37) / 10

a = 47.9181... m/s²

the crate weight is mg = 10(9.8) = 98 N.

In the vertical direction the acceleration is

a = ((600sin37 - 98) / 10)

a = 26.3089... m/s²

total acceleration is

a = √(47.9181² + 26.3089²)

a = 54.6653... m/s²

s = ½at²

t = √(2s/a)

t = √(2(1.0)/54.6653)

t = 0.19127...

t = 0.19 s