This is the second part of this question, so we need to model it as a new set of SUVAT equations. We have the initial velocity as 3.5 m/s and the final as zero. we also know the acceleration, remember to use a negative 0.5 as it is retardation.

Using the same SUVAT equations as the previous part, we can use the equation v =u +at

a simple transpose and we can solve for the time. Now we have 7 seconds as the time taken to come to rest.

There is a problem with this question. It doesn't say if the force is an elevated force (going up) or a depressive force (pushing down). We should assume that it's going up (always) as otherwise, the box will not move for part b!

First we will draw a diagram, then work on splitting that diagonal force into its Vertical and horizontal components. We will use SOH CAH TOA to split up the force. This is important because the HORIZONTAL force will be pushing (or pulling its the same) the box along the plane, and the VERTICAL force will be contributing to, or reducing the friction felt by the box.

So now we have the horizontal push, we can calculate the acceleration, the vertical "lift" doesn't affect this part of the question, as we have no friction, but we will need it for part b.

This is a very simple question. We can use F=ma to calculate the acceleration:

As this box is being pushed along a horizontal plane with no friction, all we need to solve this is a simple SUVAT:

A simple SUVAT is needed to solve this:

As there is no friction, and it is a horizontal surface, this becomes a VERY easy question, we can use F=ma to solve for the acceleration:

First, lets draw a diagram:

Next, we need to look at the vertical and horizontal forces. The force being applied is at an angle, so let's break that down, I will label the vertical y, and the horizontal x

As the angle is upwards, this is actually pulling our box up, and making the friction force less, we can use SOH CAH TOA to solve for both the horizontal and the vertical:

Ok so we have split up the diagonal force into the vertical (pulling up) and the horizontal (pushing along, or driving force). Now lets in clude gravity, which is pulling the box down.

As the pull up, and the push down are in opposite directions, we can subtract the lifting force, from the force due to gravity to find the net vertical force:

Lastly, we can calculate the coefficient of friction, as the box is "just" in motion, we know the friction force must equal to the pushing force:

It is very common to have to solve for the two unknowns in SUVAT equations, so after we have found the time, we will have to find the distance traveled, this can be done using the other SUVAT equation:

in this example my student requested I use D for distance rather than S, so I did! We are dealing with a lot of formulas, so I usually encode these as mnemonics to help us remember them. The mnemoic may be different for each person who learns it, also I work with a lot of students with dislexia, so sometimes it helps to change the letters. Whatever works!

This is a simple plug in and solve:

This is a nice easy question after such hard ones previously. A simple F=ma will do!

This is a simple suvat equation to find the acceleration, from there we will use F=ma to find the force required to create the acceleration. Let's start with a diagram and stating what we know:

We can use the second suvat equation stated to find the acceleration by transposing.

Now we know the acceleration (we know the mass is 100kg) we can calculate the force required using F=ma:

First lets draw a simple diagram. Then we need to know how much friction force is created between the box and the surface. In order to do this, we need to know the Normal force created by gravity. That is how much the mass (the weight) of the box is pushing down on the ground. The more downwards push, the more friction we will have.

The normal force (downwards force) is given by mass x 9.81

We can see that the friction created is 68.67N. Now we are pushing this box with 75N, so we need to find how much force we have left over, this will be dedicated to the acceleration.

We can see we only have 6.33N of "push" left over after we subtract the friction force. It's not much but it's something! Let's use it to do a quick F=ma calculation (again) and we now have our value for the acceleration.

First, we need to calculate how much the box is pushing down on the ground. As its being pulled down by gravity, we use F=mg (same as F=ma) to calculate the downwards force as 392.4N.

Next, we need to find the net vertical force, so we are pulling the box upwards with our applied force, so we need to subtract this:

using this vertical force, we can calculate the total amount of friction force felt on the box:

So this is the resisting force. We now need the net horizontal force, so we can take our pushing force, and subtract this frictional force. The force that is left over we will use F=ma to calculate the new acceleration: