A contestant in a winter games event pushes a 59.0 kg block of ice across a frozen lake as shown in the figure below. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03. The block is being pushed at a direction of 23 degrees below the horizontal.
(a) Calculate the minimum force F he must exert to get the block moving.
(b) What is its acceleration once it starts to move, if that force is maintained?
I am stuck when it comes to the x and y direction and how they are related. Please help!
This is wrong
To solve this problem, we can break the forces into their components in the x and y directions.
Let's start by finding the force of static friction, which is the force that opposes the motion. The formula for the force of friction is given by:
f_static = coefficient of static friction * normal force
The normal force (N) is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal to the weight of the block, which can be calculated as:
N = m * g
where m is the mass of the block and g is the acceleration due to gravity (9.8 m/s^2).
Now, let's calculate the normal force:
N = 59.0 kg * 9.8 m/s^2 = 578.2 N
The force of static friction is given by:
f_static = 0.1 * 578.2 N = 57.82 N
(a) To calculate the minimum force F required to overcome static friction and get the block moving, we need to consider the component of this force that is horizontal:
F_horizontal = F * cos(theta)
where theta is the angle below the horizontal (23 degrees in this case). Therefore:
F_horizontal = F * cos(23°)
Since the block is not moving yet, the force F must be equal to the force of static friction:
F_horizontal = f_static
Now we can solve for the minimum force F:
F * cos(23°) = 57.82 N
F = 57.82 N / cos(23°) ≈ 61.99 N
So, the minimum force F he must exert to get the block moving is approximately 62 N.
(b) Once the block starts moving, the force of static friction changes to the force of kinetic friction. The formula for the force of kinetic friction is:
f_kinetic = coefficient of kinetic friction * normal force
So, the force of kinetic friction is:
f_kinetic = 0.03 * 578.2 N ≈ 17.35 N
The force F that is being applied is constant, so the net force (F_net) in the horizontal direction is given by:
F_net = F_horizontal - f_kinetic
Now, we can use Newton's second law to calculate the acceleration (a) of the block:
F_net = m * a
Therefore:
F_horizontal - f_kinetic = m * a
a = (F_horizontal - f_kinetic) / m
Substituting the values:
a = (61.99 N - 17.35 N) / 59.0 kg ≈ 0.73 m/s^2
So, the acceleration once the block starts to move, if the force is maintained, is approximately 0.73 m/s^2.
To solve this problem, you need to break down the forces acting on the block in the x-axis and y-axis directions separately.
Let's start with the x-axis (horizontal direction). The force applied by the contestant to push the block will create a force of kinetic friction opposing the motion. The equation to calculate the force of kinetic friction is:
F_kinetic = μ_kinetic * N
where F_kinetic is the force of kinetic friction, μ_kinetic is the coefficient of kinetic friction, and N is the normal force, which is equal to the gravitational force acting on the block.
F_gravity = m * g
where F_gravity is the gravitational force, m is the mass of the block, and g is the acceleration due to gravity.
Next, let's consider the y-axis (vertical direction). The force applied by the contestant does not affect the motion in the vertical direction, as there is no vertical force component (assuming the block is on a flat surface). Therefore, the normal force acting on the block balances the gravitational force:
N = F_gravity = m * g
Now, let's calculate the minimum force required to get the block moving (part a):
First, calculate the force of kinetic friction:
F_kinetic = μ_kinetic * N = μ_kinetic * (m * g)
Then, calculate the force exerted by the contestant horizontally (in the x-axis):
F = F_kinetic
Now, to find the acceleration of the block once it starts moving (part b), you will use the force applied by the contestant:
F = m * a
where F is the force applied horizontally (minimum force to get the block moving), m is the mass, and a is the acceleration.
Rearrange the equation to solve for acceleration:
a = F / m
Substitute the value for calculated in part a into the equation above, and you'll get the acceleration of the block once it starts moving:
a = (μ_kinetic * (m * g)) / m
Simplify the equation:
a = μ_kinetic * g
Now, you can use the given values of μ_kinetic and g to find the acceleration of the block once it starts moving.
Remember to convert the angle given (23 degrees below the horizontal) into components if you need to calculate anything in the x and y direction separately.
(a)
Fnet = ma = 0
Fnet = Fapp - Ff
Fapp - Ff = 0
Fapp = Ff
F = (Fn)(μs)
F = (Fg)(0.1)
F = (9.8m)(0.1)
F = (9.8)(59)(0.1)
F = 57.82 N
(b)
Fnet = Fapp - Ff
F = 57.82 - (Fn)(μk)
F = 57.82 - (Fg)(0.03)
F = 57.82 - (9.8m)(0.03)
F = 57.82 - (9.8)(59)(0.03)
F = 57.82 - 17.346 = 40.474
F = ma = 40.474
(59)a = 40.474
a = 0.686 m/s²
The x and y directions are related because friction (a horizontal force that pushes against your push) is based on the normal force (which in this case is vertical, since the lake is a flat surface and the only force pushing down on the block is gravity).