Bodies subject to constant forces

The forces acting on the crate are its weight, the normal force of the floor and the push. The floor is smooth, therefore there is no friction and the only horizontal force is the push. The $x$ axis is chosen horizontal in the direction of the push and the $y$ axis vertical and pointing upward. Along $y$ the crate does not move, so the second law gives $N=mg$; along $x$ it gives $F=ma$, from which the acceleration $2.4m/{\mathrm{s}}^2$ follows.

The force is constant, therefore the acceleration is also constant and the motion is uniformly accelerated with zero initial velocity. The laws of uniformly accelerated motion give the velocity $9.6m/s$ and the distance traveled $19.2\mathrm{m}$. As a check, the average velocity over the $4.0\mathrm{s}$ is $\frac{0+9.6}{2}m/s=4.8m/s$, and $4.8m/s\cdot 4.0\mathrm{s}=19.2\mathrm{m}$.

The free-body diagram contains three forces: the weight, vertical and pointing downward, the normal force, vertical and pointing upward, and the force of the rope. The force of the rope is decomposed along the axes: the horizontal component $F\cos 30^{\circ }$ makes the block accelerate, the vertical component $F\sin 30^{\circ }$ pulls the block upward. The block stays on the floor, therefore the vertical acceleration is zero and the sum of the vertical components is zero.

From the vertical equation one obtains $N=27.2\mathrm{N}$, a value smaller than the weight $39.2\mathrm{N}$, because part of the weight is supported by the rope. The floor can only push, so the normal force cannot become negative, and when the vertical component of the force equals the weight one has $N=0$, and the block lifts off the floor for $F=78.5\mathrm{N}$.

The problem gives kinematic data and asks for a force, therefore one first obtains the acceleration from kinematics and then the force from the second law. The deceleration is constant, so the relation between velocity and distance of uniformly accelerated motion holds; requiring the final velocity to be zero one obtains $a=\frac{v_0^2}{2d}=6.25m/{\mathrm{s}}^2$.

Along the vertical, weight and normal force balance, and the only horizontal force is the braking force, directed opposite to the velocity. The second law gives $F=ma=7.50\cdot 10^3\mathrm{N}$. The duration of the braking is the time in which the velocity goes from $25m/s$ to zero with deceleration $6.25m/{\mathrm{s}}^2$, that is, $4.0\mathrm{s}$. As a check, the distance traveled is the average velocity times the time, $12.5m/s\cdot 4.0\mathrm{s}=50\mathrm{m}$.

The scale measures the force with which the person presses on it, which by the third law has the same magnitude as the force $N$ that the scale exerts on the person. In the reference frame of the building, which is inertial, the weight and the force $N$ act on the person; with the vertical axis pointing upward the second law gives $N=m(g+a)$, which is the formula for the apparent weight.

In the first phase the acceleration is directed upward, $a=+1.2m/{\mathrm{s}}^2$, therefore the scale reads more than the weight. In the second phase the velocity is constant, so $a=0$ and the reading coincides with the weight. In the third phase the elevator goes up and slows down, therefore the acceleration is directed downward, $a=-1.5m/{\mathrm{s}}^2$, and the scale reads less than the weight. A scale calibrated in kilograms would show about $78.6\mathrm{kg}$, $70\mathrm{kg}$ and $59.3\mathrm{kg}$. The velocity reached at the end of the first phase is $2.4m/s$, and to bring it to zero with a deceleration of $1.5m/{\mathrm{s}}^2$ takes $1.6\mathrm{s}$.