A block of wood of mass 6.0 kg slides along a skating rink at 12.5 m/s^2 [w] . The block slides onto a rough section of ice that exerts a force of 30 N force of friction on the block of wood. The acceleration of the block of wood is 5.0 m/s^2 [E] and it takes 2.5s to stop. How far does the block slide after friction begins to act on it ?
12.5 m/s^2 is not a speed. It is an acceleration rate.
What are the [w] and [E] supposed to mean?
Your 5.0 m/s^2 "acceleration" rate should be negative if the block is stopping.
Well, well, well, looks like we have a sliding block of wood here. Let's see if we can slide into an answer for you!
So, we know that the block of wood has a mass of 6.0 kg and slides along the rink at an acceleration of 12.5 m/s^2 [w], which I'm guessing stands for "wheeeee!" Oops, I mean west.
Once the block hits the rough section of ice, the force of friction comes into play and slows it down. We're informed that the acceleration becomes 5.0 m/s^2 [E], which I assume stands for "East," because the block is now frowning instead of wheeeee-ing.
Now, we need to find the distance the block slides after friction starts acting on it. To do that, we'll use the good ol' kinematic equation:
vf^2 = vi^2 + 2as
Where:
vf is the final velocity, which is 0 m/s because the block stops;
vi is the initial velocity, in this case, 12.5 m/s [w];
a is the acceleration, which is -5.0 m/s^2 because it's now going east;
and s is what we're looking for – the distance.
Plugging in the values, we get:
0 = (12.5 m/s [w])^2 + 2(-5.0 m/s^2)(s)
Now, we can solve for s:
s = -((12.5 m/s [w])^2) / (2(-5.0 m/s^2))
= -((156.25 m^2/s^2) / (-10.0 m/s^2))
= (156.25 m^2/s^2) / (10.0 m/s^2)
= 15.625 m
Ta-da! The block slides approximately 15.625 meters after friction starts acting on it. Watch out for those prankster forces, they can really slide in and cause some trouble!
To find the distance the block slides after friction begins to act on it, we can use the equations of motion:
1. Use the equation of motion to find the initial velocity of the block when friction begins to act:
v = u + at
Where:
v = final velocity (0 m/s, as the block comes to a stop)
u = initial velocity of the block (12.5 m/s)
a = acceleration of the block (5.0 m/s^2, in the opposite direction since it's slowing down)
t = time taken to stop (2.5 s)
Rearrange the equation to solve for u:
u = v - at
Substituting the given values:
u = 0 - (5.0 m/s^2)(2.5 s)
u = -12.5 m/s
Therefore, the initial velocity of the block when friction begins to act is -12.5 m/s.
2. Use the equation of motion to find the distance the block travels after friction begins to act:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity of the block (-12.5 m/s, in the opposite direction)
a = acceleration of the block (-5.0 m/s^2, opposing the motion)
t = time taken to stop (2.5 s)
Substituting the given values:
s = (-12.5 m/s)(2.5 s) + (1/2)(-5.0 m/s^2)(2.5 s)^2
Calculating:
s = -31.25 m + (-31.25 m)
s = -62.5 m
Therefore, the block slides 62.5 meters after friction begins to act on it.
To find the distance the block slides after friction begins to act on it, we need to first find the initial velocity of the block before friction acts.
Given information:
Mass of the block (m) = 6.0 kg
Acceleration of the block before friction (a1) = 12.5 m/s^2 [w]
Acceleration of the block after friction (a2) = 5.0 m/s^2 [E]
Force of friction (F) = 30 N
Time taken to stop (t) = 2.5 s
We can start by finding the initial velocity before friction using the equation:
v = u + at
where,
v = final velocity (0 m/s, as the block stops)
u = initial velocity (unknown)
a = acceleration (12.5 m/s^2 [w])
t = time taken to reach final velocity (unknown as we are finding initial velocity)
Rearranging the equation, we have:
u = v - at
Substituting the values, we get:
u = 0 - (12.5 * 2.5)
u = -31.25 m/s [w]
So, the initial velocity of the block before friction acts is -31.25 m/s [w]. Note that the negative sign indicates the direction of motion.
Next, we can find the stopping distance of the block. We can use the following equation:
v^2 = u^2 + 2as
where,
v = final velocity (0 m/s)
u = initial velocity (-31.25 m/s [w])
a = acceleration (5.0 m/s^2 [E])
s = stopping distance (unknown)
Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
Substituting the values, we get:
s = (0^2 - (-31.25)^2) / (2 * 5.0)
s = 487.19 m
Therefore, the block slides a distance of 487.19 meters after friction begins to act on it.