A man pulls a crate of mass 61.0 kg across a level floor. He pulls with a force of 186.0 N at an angle of 22.0° above the horizontal. When the crate is moving, the frictional force between the floor and the crate has a magnitude of 120.0 N. If the crate starts from rest, how fast will it be moving after the man has pulled it a distance of 2.50 m?

Question

A man pulls a crate of mass 61.0 kg across a level floor. He pulls with a force of 186.0 N at an angle of 22.0° above the horizontal. When the crate is moving, the frictional force between the floor and the crate has a magnitude of 120.0 N. If the crate starts from rest, how fast will it be moving after the man has pulled it a distance of 2.50 m?

Answer 1

Fx = 186 cos 22

Fz = 186 sin 22

186 cos 22 - 120 = 61 a

a = (1/61) 52.5
= .86 m/s^2

v = a t = .86 t

average v = .43 t

2.5 = .43 t * t

t = 2.41 seconds

so v = .86 (2.41) = 2.07 m/s

Answer 2

To find how fast the crate will be moving, we can use the work-energy principle. The work done by the man pulling the crate is equal to the change in kinetic energy of the crate.

The work done by the man is given by the formula:
Work = Force × Distance × cos(θ)

where
Force = 186.0 N (the force with which the man pulls the crate)
Distance = 2.50 m (the distance the crate is pulled)
θ = 22.0° (the angle above the horizontal)

Using this formula, we can calculate the work done by the man:
Work = 186.0 N × 2.50 m × cos(22.0°)

Now, we need to consider the work done against friction. The work done against friction is equal to the force of friction multiplied by the distance traveled:
Work against friction = Frictional force × Distance

where
Frictional force = 120.0 N (the force of friction between the floor and the crate)
Distance = 2.50 m (the distance the crate is pulled)

Now, we can calculate the net work done on the crate, which is the sum of the work done by the man and the work done against friction:
Net work = Work - Work against friction

Next, we can equate the net work done to the change in kinetic energy of the crate:
Net work = ΔKE

Since the crate starts from rest, its initial kinetic energy is zero.
ΔKE = 0.5 mv^2 - 0, where m is the mass of the crate and v is its final velocity.

Finally, we can rearrange the equation to solve for the final velocity:
v^2 = (2 × Net work) / m
v = √[(2 × Net work) / m]

The mass of the crate is given as 61.0 kg, so we can substitute the values to find the final velocity.