4 A box of mass 16 kg is on a uniformly rough horizontal floor with an applied force of fixed direction but varying magnitude \(P\) N acting as shown in Fig. 4. You may assume that the box does not tip for any value of \(P\). The coefficient of friction between the box and the floor is \(\mu\).
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\caption{Fig. 4}
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Initially the box is at rest and on the point of slipping with \(P = 58\).
- Show that \(\mu\) is about 0.25 .
In the rest of this question take \(\mu\) to be exactly 0.25 .
The applied force on the box is suddenly increased so that \(P = 70\) and the box moves against friction with the floor and another horizontal retarding force, \(S\). The box reaches a speed of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from rest after 5 seconds; during this time the box slides 3 m . - Calculate the work done by the applied force of 70 N and also the average power developed by this force over the 5 seconds.
- By considering the values of time, distance and speed, show that an object starting from rest that travels 3 m while reaching a speed of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in 5 seconds cannot be moving with constant acceleration.
The reason that the acceleration is not constant is that the retarding force \(S\) is not constant.
- Calculate the total work done by the retarding force \(S\).