# WorkA boys drags a wooden crate with a distance of 12m, across a rough level floor at a constant speed of 1.5 m/s by pulling on the rope tied to the crate with a force of 50 N. The rope makes an...

Work

A boys drags a wooden crate with a distance of 12m, across a rough level floor at a constant speed of 1.5 m/s by pulling on the rope tied to the crate with a force of 50 N. The rope makes an angle of 25 degree with the horizontal.

a. what are th horizontal and vertical components of the applied force?

b. what is the magnitude of each of the forces?

c. how much work is done by each of the forces?

d. what is the total amount of work done on the object?

e. what is coefficient of the friction of the crate on the floor?

### 2 Answers | Add Yours

The boy needs to apply a force of 50 N to drag the wooden crate across the level floor due to the force of friction that resists the motion of the crate.

Now we are given that the rope which is used to pull the crate makes an angle of 25 degrees with the horizontal.

a) The horizontal component of the applied force is 50* cos 25. The vertical component of the force applied is 50*sin 25

b) The magnitude of the horizontal component of the force is 50* cos 25 = 50* 0.9063 = 45.315 N. The magnitude of the vertical component of the force is 50* sin 25 = 50*0.4226 = 21.13 N.

c) As the crate moves only in the horizontal direction and does not have any vertical displacement only the horizontal component of the force applied does work. No work is done by the vertical component of the force applied.

d) The total work done on the object is the product of the horizontal displacement and the horizontal component of the force applied. This is 12 * 45.315 = 543.78 J.

e) When the horizontal component of the force applied is equal to the force of kinetic friction the crate will move at a constant speed. Here the vertical component of the applied force is 21.13. The force of the kinetic friction is equal to Ff*[m*9.8 - 21.13], where Ff is the coefficient of friction, m is the mass of the crate and 9.8 is the acceleration due to gravity. This has to be equal to the horizontal component of the force applied as the crate moves at a constant speed. So we have Ff*[m*9.8 - 21.13] = 45.315. Here as we are not given the mass of the crate we cannot determine the exact value of the coefficient of friction.

The figure is below.

a)

The force applied can be decomposed into its horizontal and vertical components.

`F_h = F*cos(alpha) =50*cos(25) =45.32 N`

`F_v = F*sin(alpha) = 50*sin(25) =21.13 N`

b)

The speed of the box is constant, it means there is no acceleration. The sum of forces on the horizontal and vertical axis need to be zero.

On the horizontal: horizontal component of force need to equal the friction force `F_h = Ff =45.32 N`

On the vertical: the difference between the box weight and the vertical force is equal to the reaction from the ground`G-Fv =N`

Since the mass of the box is not give one can not determine the magnitude of these forces on the vertical axis.

c) On the horizontal axis the work done by the horizontal force (positive) is equal in absolute value with the work done by the friction force (negative)

`W = F_h*d =45.32*12 =543.84 J`

On the vertical axis because there is no displacement the work done by each of the three forces is zero.

d) As stated above, the total amount of work done by all forces on the object is zero.

e) Since the mass of the object is not given, the coefficient of friction can not be determined.