A 3.1 kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 16.6 N. Find the speed after it has moved 3.8m.
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you are given:
u: 0
s: 3.8
F: 16.6
m: 3.1
As F=ma, a=F/m, making a=16.6/3.1=5.35483871
kinematics equation:
v^2=u^2+2as
v^2=0^2+2(5.35483871)(3.8)
v^2=40.6967742
v=6.379402339 m/s
Why did the block bring a tape measure to the party? Because it wanted to measure its speed after moving 3.8 meters!
To find the speed, we can use the equation:
v = √(2ad)
Where:
v = final speed
a = acceleration
d = displacement
In this case, the block is pulled by a constant force, so the acceleration can be found using Newton's second law:
F = ma
Where:
F = force applied
m = mass of the block
a = acceleration
Solving for a:
a = F/m
Substituting the values given:
a = 16.6 N / 3.1 kg
Now we can substitute the values for acceleration and displacement into the equation for speed:
v = √(2 * a * d)
v = √(2 * (16.6 N / 3.1 kg) * 3.8 m)
v = √(2 * 10.3 N/kg * 3.8 m)
v = √(78.44 N*m/kg)
v ≈ 8.85 m/s
So, the speed of the block after moving 3.8 meters is approximately 8.85 m/s.
To find the speed of the block after it has moved 3.8m, we can use the principles of Newton's second law.
Step 1: Determine the acceleration of the block.
Since the block is initially at rest, we can assume that the initial velocity, u = 0. Using Newton's second law, F = ma, we can rearrange the equation to solve for acceleration:
a = F/m,
where F is the force applied to the block and m is the mass of the block.
Given:
Force, F = 16.6 N,
Mass, m = 3.1 kg.
Plugging the values into the equation:
a = 16.6 N / 3.1 kg,
a ≈ 5.35 m/s².
Step 2: Use kinematic equations to find the final velocity.
Let's assume the final velocity is v.
We know the initial velocity, u = 0, the acceleration, a = 5.35 m/s², and the displacement, s = 3.8 m.
The kinematic equation relating initial velocity, final velocity, acceleration, and displacement is:
v² = u² + 2as.
Plugging in the values:
v² = 0² + 2 * 5.35 m/s² * 3.8 m,
v² = 40.632 m²/s².
Taking the square root of both sides to solve for v:
v = √40.632 m²/s²,
v ≈ 6.37 m/s.
Therefore, the speed of the block after it has moved 3.8m is approximately 6.37 m/s.
a = F/m = 16.6/3.1
d = (1/2) a t^2
3.8 = (8.3/3.1) t^2
t = 1.19 seconds
v = a t = (16.6/3.1)(1.19)
= 6.38 m/s
or do using change in Ke = work done
(1/2) 3.1 v^2 = 16.6*3.8
v = 6.38 m/s
a = F/m = 16.6/3.1
d = (1/2) a t^2
3.8 = (8.3/3.1) t^2
t = 2.01 seconds
v = a t = (16.6/3.1)(2.01)