Part 1

A block of mass 4 kg, which has an initial
speed of 8 m/s at time t = 0, slides on a
horizontal surface.Find the magnitude of the work that must
be done on the block to bring it to rest.
Answer in units of J.

Part 2
If a constant friction force of magnitude
5 Newtons is exerted on the block by the surface, find the magnitude of the acceleration of
the block.
Answer in units of m/s2.

Part 3
How far does the block slide before it comes
to rest?
Answer in units of m.

1.

enough work must be done to reduce the Ke to zero
so
work = (1/2) m v^2 Joules

2.
a= F/m = 5/4 m/s^2

3.
work done by friction = initial Ke when it stops
5 * d = (1/2) m v^2

Part 1: To find the magnitude of the work done on the block to bring it to rest, we need to use the concept of work and energy. The work done on an object is equal to the change in its kinetic energy. In this case, the initial kinetic energy is given by (1/2)mv^2, where m is the mass of the block (4 kg) and v is the initial speed (8 m/s). Since the block comes to rest, its final kinetic energy is zero.

Therefore, the work done to bring the block to rest is equal to the initial kinetic energy:

Work = (1/2)mv^2 = (1/2)(4 kg)(8 m/s)^2 = 128 J.

So, the magnitude of the work done on the block to bring it to rest is 128 J.

Part 2: To find the magnitude of the acceleration of the block, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the friction force exerted on the block, and its magnitude is given as 5 N. The mass of the block is 4 kg.

Therefore, we can write the equation as:
5 N = 4 kg * a,

where a is the acceleration.

To find the magnitude of the acceleration, we rearrange the equation:

a = (5 N) / (4 kg) = 1.25 m/s^2.

So, the magnitude of the acceleration of the block is 1.25 m/s^2.

Part 3: To find how far the block slides before it comes to rest, we can use the equation of motion that relates distance, initial velocity, acceleration, and time. The block starts with an initial velocity of 8 m/s and comes to rest, so its final velocity is 0 m/s. We need to find the distance traveled.

The equation that relates the variables is:
v^2 = u^2 + 2as,

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

Rearranging the equation, we have:
s = (v^2 - u^2) / (2a).

Substituting the values, we get:
s = (0^2 - 8^2) / (2 * -1.25 m/s^2) = (-64) / (-2.5) = 25.6 m.

So, the block slides a distance of 25.6 m before it comes to rest.