Question 5.1a The figure shows a box of mass m-4.55 kg puiled to the right acros

Question

Question 5.1a The figure shows a box of mass m-4.55 kg puiled to the right across a horizontal surface by a constant tension force of magnitude T- 29.5 N. The tension force is inclined at an angle 0-24.0° above the horizontal and the friction force has magnitude 12.5 N. The box is puled a distance d-1.75 m. Determine the work done on the box by the tension force, the normal force, and the kinetic friction force. We'l use W War and Wa to represent the magnitudes of the normal, and friction forces, respectively WT

Explanation / Answer

Given,

m = 4.55 kg ; T = 29.5 N ; theta = 24 deg ; Ff = 12.5 N

The work done by the tension force is:

Wt = T d cos(theta)

Wt = 29.5 x 1.75 x cos24 = 47.16 J

Hence, Wt = 47.16 J

The work done by the normal force is zero.

Wn = 0 J

The work done by the frictional force is:

Wf = Ff d cos(theta)

Wf = 12.5 x 1.75 x cos24 = 19.98 J

Hence, Wf = 19.98 J

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