odetteerivera
26.09.2019 •
Physics
Aboy 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?
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Ответ:
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
Ответ:
To calculate the value of unknown charge, we use the formula the electric force between two charges
Here, q is unknown charge , , and
Now substituting the given values in above formula we get
or
The unknown charge is positive because the force between charges is attractive.