Question

A boy pulls a 30.0-kg box with a 210-N force at 39° above a horizontal surface. If the coefficient of kinetic friction between the box and the horizontal surface is is 0.23 and the box is pulled a distance of 38.0 m, what is the work done by the friction force on the box?

Answer 1

Answer:

W= - 1414.48 J  : Work done by the friction force on the box.

Explanation:

Work concept

The work (W) of a given force is defined as the product of the force (F) by the distance(d) in which the force acted.

W= F*d :in (Joules)  Formula (1)

Friction force concept

Ff=μk*N  Formula (2)

μk=coefficient of kinetic friction

N= Normal force

We apply Newton's first law for forces in the direction of the axis y

∑Fy=0 Equation (1)

The forces acting on the box in the direction of the axis and are the following:

• Fy=Force of 210 N in y= 210*sen 39°= 132.16 N ( +y)

• W=Weight of box= m*g=30kg*9.8m/s² = 294N (-y)
• N =Normal force ( +y)

We apply equation (1) and do the algebraic sum of forces in y

Fy+N-W=0

132.16+N-294=0

N= 294-132.16=161.84 N

We apply formula (2) to calculate Ff:

Ff= -0.23*161.84=-37,22N  The frictional force is negative because it goes in the opposite direction to the displacement.

We apply formula (1) to calculate the work done by the friction force on the box

W= - 37,22 N * 38 m = -1414.48 N*m = -1414.48 J