a) The coefficient of kinetic friction between a steel block and an ice rink surface is 0.0100. If a force of 24.5 N keeps the steel block moving at a steady speed, what is the force of gravity on the block?
b)A copper block has dimensions 1 cm × 2 cm × 4 cm. A force of 0.10 N will pull the block along a table surface at a steady speed if the 1 cm × 4 cm side is face down on the table. What force will be needed to pull the same block along when its 2 cm × 4 cm side is face down?
a) Ah, the age-old battle between friction and gravity! To find the force of gravity on the block, we need to look at the steady speed scenario. In this case, the force of friction counteracts the force applied to keep the block moving. So, we can use the equation:
Force of friction = coefficient of kinetic friction * Force of gravity
Since the block is moving at a steady speed, the force of friction is equal to the force keeping the block moving, which is 24.5 N. Now, let's rearrange the equation to solve for the force of gravity:
Force of gravity = Force of friction / coefficient of kinetic friction
Plugging in the given values, we get:
Force of gravity = 24.5 N / 0.0100
So, the force of gravity on the block is approximately 2450 N, assuming my calculations aren't slippier than the block itself!
b) Copper block, you've got some dimensions! But fear not, I will help you calculate the force needed to pull you along when your side changes.
Let's first deal with the side facing down on the table. Since the block moves at a steady speed, the force of friction will be equal to the applied force. Using the equation:
Force of friction = coefficient of kinetic friction * Force acting on the block
We know that the force of friction is 0.1 N (because it's equal to the pulling force), and we need to determine the force acting on the block.
Now the tricky part! When the 1 cm × 4 cm side is face down on the table, the force acting on the block is the force applied parallel to the surface. So, we need to calculate the area of that side:
Area = length × width = 1 cm × 4 cm = 4 cm²
Now, to figure out the force needed when the 2 cm × 4 cm side is face down, we need to find the force acting on the block in that orientation. So, let's calculate its area:
Area = length × width = 2 cm × 4 cm = 8 cm²
Since the force of friction is the same in both cases, we can use the formula:
Force of friction = coefficient of kinetic friction * Force acting on the block
For the new orientation, we need to determine the force acting on the block with the larger area:
Force acting on the block = Force of friction / coefficient of kinetic friction
Plugging in the given values, we get:
Force acting on the block = 0.1 N / coefficient of kinetic friction
To be honest, the coefficient of kinetic friction was not given, so I can't give you an exact answer. So, let's just assume it has a super smooth coefficient of 0.5 for simplicity:
Force acting on the block = 0.1 N / 0.5
Thus, the force needed to pull the same block along when the 2 cm × 4 cm side is face down would be approximately 0.2 N. Remember, this is just for a fictional coefficient. If you have the correct coefficient of kinetic friction, feel free to plug it in and calculate accurately!
a) To find the force of gravity on the steel block, we need to consider the equilibrium of forces acting on the block.
The force of gravity (F_gravity) on the block can be calculated using the equation:
F_gravity = F_applied - F_friction,
where F_applied is the applied force and F_friction is the force of friction.
Given:
Coefficient of kinetic friction (μ_kinetic) = 0.0100
Applied Force (F_applied) = 24.5 N
The force of friction (F_friction) can be calculated using the formula:
F_friction = μ_kinetic * N,
where N is the normal force.
Since the block is not accelerating, the normal force (N) is equal to the force of gravity (F_gravity).
Substituting the values into the formula, we have:
F_friction = 0.0100 * F_gravity.
Now, we can rewrite the equation for force of gravity as:
F_gravity = F_applied - F_friction.
Substituting the values, we get:
F_gravity = 24.5 N - (0.0100 * F_gravity).
Simplifying the equation, we find:
F_gravity + 0.0100 * F_gravity = 24.5 N.
Combining like terms, we have:
1.0100 * F_gravity = 24.5 N.
Dividing both sides by 1.0100:
F_gravity = 24.5 N / 1.0100.
Calculating the value:
F_gravity = 24.26 N.
Therefore, the force of gravity on the block is approximately 24.26 N.
b) To find the force required to pull the copper block along when its 2 cm × 4 cm side is face down, we need to consider the equilibrium of forces.
Given:
Applied Force (F_applied) = 0.10 N
When the 1 cm × 4 cm side is face down, only a fraction of the force is needed to maintain a steady speed due to the smaller contact area.
To calculate the force needed when the 2 cm × 4 cm side is face down, we can use the concept of pressure, which is force per unit area.
The pressure exerted by the block on the table can be expressed as:
Pressure = Force / Area.
When the 1 cm × 4 cm side is face down, the contact area is 1 cm × 4 cm = 4 cm^2.
Therefore, the pressure is:
Pressure1 = F_applied / (1 cm × 4 cm).
When the 2 cm × 4 cm side is face down, the contact area is 2 cm × 4 cm = 8 cm^2.
To calculate the force needed, we can use the equation:
Force2 = Pressure1 × Area2,
where Area2 is the new contact area.
Substituting the values, we have:
Force2 = Pressure1 × (2 cm × 4 cm).
Calculating the value:
Force2 = (0.10 N) / (1 cm × 4 cm) × (2 cm × 4 cm).
Simplifying the equation, we find:
Force2 = 0.10 N × (8 cm^2 / 4 cm^2).
Force2 = 0.10 N × 2.
Force2 = 0.20 N.
Therefore, the force needed to pull the same block along when its 2 cm × 4 cm side is face down is 0.20 N.
a) To find the force of gravity on the block, we need to first understand the concept of kinetic friction. Kinetic friction is the force that opposes the motion of an object when it is in contact with a surface.
The formula for kinetic friction is given by:
\( f_{k} = \mu_{k} \cdot N \)
Where:
\( f_{k} \) is the force of kinetic friction
\( \mu_{k} \) is the coefficient of kinetic friction between the block and the ice rink surface
\( N \) is the normal force exerted on the block
In this case, we know the coefficient of kinetic friction (\( \mu_{k} \)) is 0.0100 and the force of kinetic friction (\( f_{k} \)) is given as 24.5 N. We need to find the force of gravity on the block.
Now, since the block is moving at a steady speed, we know that the force of friction (\( f_{k} \)) is equal and opposite to the applied force (24.5 N). Therefore, we can equate the force of kinetic friction to the applied force:
\( f_{k} = 24.5 \) N
Next, we can rearrange the formula for kinetic friction to solve for the normal force (\( N \)):
\( N = \frac{{f_{k}}}{{\mu_{k}}} \)
Substituting the given values, we have:
\( N = \frac{{24.5}}{{0.0100}} \) N
Calculating this, we find:
\( N = 2450 \) N
The normal force (\( N \)) is equal to the force of gravity acting on the block when no other vertical forces are present. Therefore, the force of gravity on the block is also 2450 N.
b) To find the force needed to pull the copper block along when its 2 cm × 4 cm side is face down, we need to consider the relationship between force, friction, and area of contact.
The force of friction (\( f_{k} \)) can be calculated using the formula mentioned above:
\( f_{k} = \mu_{k} \cdot N \)
Where:
\( \mu_{k} \) is the coefficient of kinetic friction
\( N \) is the normal force
Since the coefficient of kinetic friction and the material (copper) remain the same, the only difference is the area of contact. When the 1 cm × 4 cm side is face down, the area of contact is \( 1 \, \text{cm} \times 4 \, \text{cm} = 4 \, \text{cm}^2 \).
Now, let's calculate the force of friction (\( f_{k} \)) using the given force of 0.10 N:
\( 0.10 \, \text{N} = \mu_{k} \cdot N \)
We need to find the normal force (\( N \)). Using the area of contact (4 cm^2) as the surface area, we can calculate it:
\( N = \frac{{0.10 \, \text{N}}}{{\mu_{k}}} \)
Next, we need to find the new force of friction when the 2 cm × 4 cm side is face down. The area of contact in this case is \( 2 \, \text{cm} \times 4 \, \text{cm} = 8 \, \text{cm}^2 \).
Using the same formula for kinetic friction:
\( f_{k} = \mu_{k} \cdot N \)
Substituting the new normal force (\( N' \)) for the 2 cm × 4 cm side:
\( f_{k}' = \mu_{k} \cdot N' \)
Now, since the force is being applied to move the block at a steady speed, the force of friction will be the same (0.10 N). Therefore, we can set up the following equation:
\( \mu_{k} \cdot N' = 0.10 \, \text{N} \)
To find the force needed to pull the block along when the 2 cm × 4 cm side is face down, we need to solve for \( N' \):
\( N' = \frac{{0.10 \, \text{N}}}{{\mu_{k}}} \)
Substituting the given coefficient of kinetic friction, we can calculate the force needed:
\( N' = \frac{{0.10 \, \text{N}}}{{\mu_{k}}} \)