A block of mass 4 kg, which has an initial
speed of 2 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.
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/s^2
How far does the block slide before it comes
to rest?
Answer in units of m
Woah
Kinetic energy = (1/2) m v^2
= (1/2)(4)(4)
= 8 joules
so you need 8 Joules to stop it
F = m a
5 = 4 a
a = 5/4 m/s^2 = 1.25 m/s^2
5 x = 8
x = 8/5 = 1.6 meter
(1/2)(4)(2)(2) is the same as (1/2)(4)(4)
Why did the block go to therapy? Because it couldn't stop sliding and needed to work through its acceleration issues! Let's solve these problems one by one, shall we?
1. To find the magnitude of the work done on the block to bring it to rest, we'll use the work-energy principle. The work done on an object is equal to its change in kinetic energy. Since the block comes to rest, its final kinetic energy is zero. Therefore, the work done on the block is equal to its initial kinetic energy. The formula for kinetic energy is KE = 1/2 * mass * velocity^2. Plugging in the values, we get KE = 1/2 * 4 kg * (2 m/s)^2 = 8 J. So, the magnitude of the work done on the block is 8 Joules.
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 acting on the block is the force of friction. Therefore, we have F = m * a. Rearranging the equation, we have a = F/m = 5 N / 4 kg = 1.25 m/s^2. So, the magnitude of the acceleration of the block is 1.25 m/s^2.
3. To find the distance the block slides before it comes to rest, we can use the equation for motion under constant acceleration: d = (v^2 - u^2) / (2a), where d is the distance, v is the final velocity (which is zero in this case), u is the initial velocity, and a is the acceleration. Plugging in the values, we get d = (0 - (2 m/s)^2) / (2 * 1.25 m/s^2) = -4 / 2.5 = -1.6 m. Since distance can't be negative, we take the magnitude and the block slides a distance of 1.6 meters before coming to rest.
So, the block reluctantly slid 1.6 meters, and the work done on it was 8 Joules. I hope that brings a smile to your face!
To find the magnitude of the work done on the block to bring it to rest, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
First, we need to find the initial kinetic energy of the block. The kinetic energy (KE) is given by the formula KE = (1/2) * mass * velocity^2.
Given:
Mass (m) = 4 kg
Initial velocity (v) = 2 m/s
Initial kinetic energy (KE) = (1/2) * 4 kg * (2 m/s)^2
= 8 J
Since the block comes to rest, its final kinetic energy will be zero. Therefore, the work done on the block to bring it to rest is the negative of the initial kinetic energy.
Magnitude of work done to bring the block to rest = -8 J
To find the magnitude of the acceleration of the block, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is the frictional force.
Given:
Frictional force (F) = 5 N
Mass (m) = 4 kg
Using the formula F = m * a, we can rearrange the equation to find the acceleration (a).
a = F / m
= 5 N / 4 kg
= 1.25 m/s^2
Magnitude of the acceleration of the block = 1.25 m/s^2
To find the distance the block slides before it comes to rest, we can use the equation of motion that relates distance, initial velocity, acceleration, and time. In this case, we need to find the time it takes for the block to come to rest, and then use that time to calculate the distance.
The equation for time is given by:
vf = vi + a * t
Since the final velocity (vf) is zero when the block comes to rest, we can rearrange the equation to solve for time:
t = -vi / a
= -2 m/s / -1.25 m/s^2
= 1.6 s
Now, using the equation for distance:
d = vi * t + (1/2) * a * t^2
Plugging in the values:
d = 2 m/s * 1.6 s + (1/2) * -1.25 m/s^2 * (1.6 s)^2
= 3.2 m + (1/2) * -1.25 m/s^2 * 2.56 s^2
= 3.2 m - 1.6 m
= 1.6 m
Therefore, the block slides a distance of 1.6 meters before it comes to rest.
First part.
=(1/2)(4)(2)(2)