A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 295 N a

Question

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 295 N at a speed of v = 0.860 m/s across the warehouse floor. He encounters a rough horizontal section of the floor that is 0.75 m long and where the coefficient of kinetic friction between the crate and floor is 0.360.

(a) Determine the magnitude and direction of the net force acting on the crate while it is pushed over the rough section of the floor.

direction: up, down, same direction, opposite direction of motion of crate

(b) Determine the net work done on the crate while it is pushed over the rough section of the floor.

____ J

(c) Find the speed of the crate when it reaches the end of the rough surface.

____ m/s

magnitude ____N

direction: up, down, same direction, opposite direction of motion of crate

(b) Determine the net work done on the crate while it is pushed over the rough section of the floor.

____ J

(c) Find the speed of the crate when it reaches the end of the rough surface.

____ m/s

A warehouse worker is pushing a 90.0-kg crate with a horizontal force of 295 N at a speed of v = 0.860 m/s across the warehouse floor. He encounters a rough horizontal section of the floor that is 0.75 m long and where the coefficient of kinetic friction between the crate and floor is 0.360. (a) Determine the magnitude and direction of the net force acting on the crate while it is pushed over the rough section of the floor. direction: up, down, same direction, opposite direction of motion of crate (b) Determine the net work done on the crate while it is pushed over the rough section of the floor. (c) Find the speed of the crate when it reaches the end of the rough surface.

Explanation / Answer

a) Kinetic Friction Force f = u.N

and N = mg

so f = 0.360 x mg = 0.360 x 90 x 9.81 = 317.52 N

So Fnet = F - f = 295 - 317.52 = - 22.52 N

Magnitude = 22.52 N

Direction = Opposite diretion of motion

B) Net Work Done = Fnet .d

= -22.52 * 0.75   = - 16.89 J

C ) Using Work energy theorem,

Net work Done= change in K.E.

-16.89 = 90v^2 /2 - 90*0.860^2 /2

v = 0.604 m/s ...Ans

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