A boy pulls a bag of baseball bats across a ball field toward the parking lot. The bag of bats has a mass of 6.8 kg, and the boy exerts a horizontal force of 20 N on the bag. As a result, the bag accelerates from rest to a speed of 1.12 m/s in a distance of 5.4 m.
Part A
What is the coefficient of kinetic friction between the bag and the ground? 3sf
V^2 = Vo^2 + 2a*d.
1.12^2 = 0 + 2a*5.4, a = 0.116 m/s^2.
M*g = 6.8*9.8 = 66.64 N. = Wt. of bats. = Normal force(Fn).
Fk = u*Fn = 66.64u.
Fe-Fk = M*a.
20-66.64u = 6.8*0.116 = 0.789,
66.64u = 20-0.789 = 19.2,
u = 0.288.
Oh, coefficient of kinetic friction, huh? Time to put on my physics hat and give you an answer with a touch of humor!
Alright, let's break it down. The boy is pulling the bag of baseball bats across the ball field, so we're dealing with some friction here. Now, the equation for kinetic friction is Fk = μk * N, where Fk is the force of kinetic friction, μk is the coefficient of kinetic friction, and N is the normal force.
We know that the boy exerts a horizontal force of 20 N on the bag. So, we can say that Fk = 20 N.
Now, let's find the normal force. Since the bag is on the ground, the normal force will be equal to the weight of the bag, which is given by N = m * g, where m is the mass of the bag (6.8 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we get N = 6.8 kg * 9.8 m/s^2. So N is approximately 66.64 N.
Now, let's rearrange the equation for kinetic friction to solve for the coefficient of kinetic friction, μk = Fk / N. Plugging in the values we know, we get μk = 20 N / 66.64 N.
After crunching the numbers, we find that the coefficient of kinetic friction between the bag and the ground is approximately 0.30.
So, the coefficient of kinetic friction between the bag and the ground is like that one friend who can't resist grabbing on to your leg when you're trying to walk away. It's always causing a little resistance, making you work harder to drag that bag of bats across the field!
To find the coefficient of kinetic friction between the bag and the ground, we can use the equation:
ma = μk * m * g
Where:
m = mass of the bag (6.8 kg)
a = acceleration of the bag (1.12 m/s^2)
μk = coefficient of kinetic friction
g = acceleration due to gravity (9.8 m/s^2)
First, let's calculate the net force acting on the bag:
Fnet = ma
Fnet = m * a
Fnet = 6.8 kg * 1.12 m/s^2
Fnet = 7.616 N
Next, let's calculate the force of kinetic friction using the equation:
Ff = μk * m * g
Ff = μk * (6.8 kg) * (9.8 m/s^2)
Ff = μk * 66.64 N
Now, we can equate the net force and the force of kinetic friction:
Fnet = Ff
7.616 N = μk * 66.64 N
Solving for μk:
μk = 7.616 N / 66.64 N
μk = 0.1144
Rounded to three significant figures, the coefficient of kinetic friction between the bag and the ground is approximately 0.114.
To find the coefficient of kinetic friction between the bag and the ground, we can use Newton's second law and the equation for friction.
The first step is to calculate the net force acting on the bag. We know that the boy exerts a horizontal force of 20 N. The net force can be found using Newton's second law:
Net force = mass × acceleration
Given:
Mass of the bag (m) = 6.8 kg
Acceleration of the bag (a) = (final velocity - initial velocity) / distance
Final velocity (v) = 1.12 m/s
Initial velocity (u) = 0 m/s (the bag starts from rest)
Distance (d) = 5.4 m
Plugging in the values:
Acceleration (a) = (1.12 m/s - 0 m/s) / 5.4 m
= 1.12 m/s / 5.4 m
= 0.2074 m/s^2
Net force = 6.8 kg × 0.2074 m/s^2
= 1.40712 N
Now, we need to find the force of kinetic friction (Fk) using the equation for friction:
Fk = coefficient of kinetic friction × normal force
The normal force acting on the bag is equal to the weight of the bag, which is given by:
Normal force = mass × gravitational acceleration
Mass of the bag (m) = 6.8 kg
Gravitational acceleration (g) = 9.8 m/s^2
Normal force = 6.8 kg × 9.8 m/s^2
= 66.64 N
Now, rearrange the equation for friction to solve for the coefficient of kinetic friction:
Coefficient of kinetic friction = Fk / Normal force
Coefficient of kinetic friction = 1.40712 N / 66.64 N
= 0.0211 (rounded to 3 significant figures)
Therefore, the coefficient of kinetic friction between the bag and the ground is approximately 0.0211.