Question

Question 5.1a:
The figure shows a box of mass

m = 5.10 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

? = 16.0

Answer 1

So, for the first question, the normal force and the gravity force do no work because they are perpendicular to the motion of the box. Now, the friction force acts opposite the direction of motion, so its work will be negative (friction forces always create negative work). So the work done by friction is simply the magnitude of the force multiplied by the distance traveled: W_f=(-14.0N)(1.95m). Now the tension also does work, but only the x-component of the tension. To find the x-component of the tension, all you have to do is take the magnitude of the tensional force and multiply it by the cosine of the angle it makes with the horizontal. So, T_x=(21.0N)(cos(16)). Now that you have the x-component of the tension, just multiply that value by the distance traveled. This work will be positive since the force is in the direction of motion. So, W_T=(21.0N)(cos(16))(1.95m). Thats all you have to do for the first one!

Now, for the second question, you are basically doing the same thing. The only difference is that you will want to choose the direction down the incline as your positive x direction. So for this question, all you have to do is find the component of the gravity force that acts down the incline. This is simply the magnitude of the gravity force multiplied by the sine of the angle it makes with the incline, which is the same as the angle the incline makes with the horizontal, 27 degrees. So the component of gravity down the incline is just (2.2X10⁴N)(sin(27)). Now that you have the component of gravity acting parallel to the incline, you simply multiply this value by the distance traveled to find the work done. So, W_g=(2.2X10⁴N)(sin(27))(14.5m)

There ya go, hope that helps! If so please leave me a rating!