The figure shows a box of mass m = 4.50 kg pulled to the right across a horizont

ID: 1780058 • Letter: T

Question

The figure shows a box of mass

m = 4.50 kg

pulled to the right across a horizontal surface by a constant tension force of magnitude

T = 21.0 N.

The tension force is inclined at an angle

= 15.5°

above the horizontal and the friction force has magnitude

fk = 14.5 N.

The box is pulled a distance

d = 2.15 m.

Determine the work done on the box by the tension force, the normal force, and the kinetic friction force. We'll use

WT,

Wn,

and

Wfk

to represent the magnitudes of the tension, normal, and friction forces, respectively.

Wfk

= J

Wfk

= J

here,

m = 4.5 kg

T = 21 N

theta = 15.5 degree

fk = 14.5 N

d = 2.15 m

the work done on the box by the tenion force , W = T * d * cos(theta)

W = 21 * 2.15 * cos(15.5) J

W = 43.5 J

the work done by normal force , Wn = N * d * sin(90) = 0 J

the work done by kinetic friction force , Wff = fk * d * cos(180) J

Wff = 14.5 * 2.15 * (-1) J

Wff = - 31.18 J

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