Block resting on the back of a decelerating truck

Question

During braking, a truck has a steady deceleration of 7.0 m/s^2. A box sits on the platform of this truck. The box begins to slide when the braking begins, and after sliding a distance of 2.0 m (relative to the truck), it hits the front of the truck. With what speed (relative to the truck) does the box hit? The coefficient of kinetic friction for the box is 0.50.

Conceptual Issue: In this problem, we can see that the truck is decelerating at 7.0 m/s^2. Does this mean that the box resting on the back of the truck starts moving forward towards the front of the truck with an acceleration of 7.0 m/s^2?

Answer 1

No, just because the truck has some acceleration relative to the ground does not mean that the box has the same magnitude of acceleration relative to the ground or truck (you don't specify which in your question). This is because there is friction involved here, so you need to consider that force as well.

Answer 2

I would approach this from the fact that the distance $d$ travelled by the box relative to the ground can be expressed as:

$$d_\text{box relative to ground, B/G} = d_\text{box relative to truck, B/T} + d_\text{truck relative to ground, T/G} $$

which I then rewrite as:

$$d_\text{B/G}(t) = d_{\rm B/T}(t) + d_{\rm T/G}(t) \tag{1}$$

Which leads to the following relationship (either by using kinematic equations or by taking derivative w.r.t. $t$ since Eq. (1) holds for all $t$):

$$a_{\rm B/G} = a_\text{B/T} + a_\text{T/G}. \tag{2}$$

To find $a_{\rm B/G}$, draw a Free Body Diagram to identify the forces acting on the box relative to the ground.

After that, you'll be able to solve for the acceleration of the box with respect to the truck, $a_{\rm B/T}$. Do note that relative to the truck, the initial speed of the box is $0$.

Finally, recall the kinematic equation relating distance and speed: $$v_f^2 = v_0^2+2a\Delta d.$$

Use that equation in the (non inertial) reference frame that is moving with the truck (such that the $d = d_{\rm B/T} = 2 \text{ m}$ and $a=a_{\rm B/T}$.

This should be enough information to solve it.