Question

1. in figure 4.7, the net external force on the 24-kg mower is stated to be 51 n. if the force of friction opposing the motion is 24 n, what force f (in newtons) is the person exerting on the mower? suppose the mower is moving at 1.5 m/s when the force f is removed. how far will the mower go before stopping?

Explanation:

Part 1: Finding the force \( F \) exerted by the person

We know that the net force \( F_{\text{net}} \) is the vector sum of the applied force \( F \) and the frictional force \( f \) (which opposes the motion, so it's in the opposite direction of \( F \)). Assuming the motion is in the direction of \( F \), the net force formula is \( F_{\text{net}} = F - f \) (since friction acts against the motion). We need to solve for \( F \).

Step 1: Identify the known values

We know \( F_{\text{net}} = 51 \, \text{N} \) and \( f = 24 \, \text{N} \).

Step 2: Rearrange the net force formula to solve for \( F \)

From \( F_{\text{net}} = F - f \), we can rearrange it to \( F = F_{\text{net}} + f \).

Step 3: Substitute the known values into the formula

Substituting \( F_{\text{net}} = 51 \, \text{N} \) and \( f = 24 \, \text{N} \) into the formula, we get:
\( F = 51 + 24 = 75 \, \text{N} \)

Part 2: Finding the distance the mower travels after \( F \) is removed

When the force \( F \) is removed, the only horizontal force acting on the mower is the frictional force \( f = 24 \, \text{N} \) (opposing the motion). We can use Newton's second law to find the acceleration, then use kinematic equations to find the distance.

Step 1: Find the acceleration using Newton's second law

Newton's second law is \( F = ma \), where \( F \) here is the net force (which is the frictional force, but since it's opposing the motion, the acceleration will be negative, indicating deceleration). So \( -f = ma \) (the negative sign indicates the acceleration is in the opposite direction of motion). We know \( m = 24 \, \text{kg} \) and \( f = 24 \, \text{N} \).

Rearranging for \( a \): \( a = \frac{-f}{m} \)

Substituting the values: \( a = \frac{-24}{24} = -1 \, \text{m/s}^2 \)

Step 2: Use the kinematic equation to find the distance

The kinematic equation that relates initial velocity \( u \), final velocity \( v \), acceleration \( a \), and distance \( s \) is \( v^2 = u^2 + 2as \). We know the mower comes to a stop, so \( v = 0 \, \text{m/s} \), the initial velocity \( u = 1.5 \, \text{m/s} \), and the acceleration \( a = -1 \, \text{m/s}^2 \). We need to solve for \( s \).

Rearranging the kinematic equation for \( s \): \( s = \frac{v^2 - u^2}{2a} \)

Step 3: Substitute the known values into the kinematic equation

Substituting \( v = 0 \), \( u = 1.5 \, \text{m/s} \), and \( a = -1 \, \text{m/s}^2 \) into the formula:

\( s = \frac{0 - (1.5)^2}{2(-1)} = \frac{-2.25}{-2} = 1.125 \, \text{m} \)

Final Answers

• The force \( F \) exerted by the person is \( \boldsymbol{75 \, \text{N}} \).
• The distance the mower travels before stopping is \( \boldsymbol{1.125 \, \text{m}} \) (or \( 1.13 \, \text{m} \) if rounded to two decimal places).

Answer:

When the force \( F \) is removed, the only horizontal force acting on the mower is the frictional force \( f = 24 \, \text{N} \) (opposing the motion). We can use Newton's second law to find the acceleration, then use kinematic equations to find the distance.

Step 1: Find the acceleration using Newton's second law

Newton's second law is \( F = ma \), where \( F \) here is the net force (which is the frictional force, but since it's opposing the motion, the acceleration will be negative, indicating deceleration). So \( -f = ma \) (the negative sign indicates the acceleration is in the opposite direction of motion). We know \( m = 24 \, \text{kg} \) and \( f = 24 \, \text{N} \).

Rearranging for \( a \): \( a = \frac{-f}{m} \)

Substituting the values: \( a = \frac{-24}{24} = -1 \, \text{m/s}^2 \)

Step 2: Use the kinematic equation to find the distance

The kinematic equation that relates initial velocity \( u \), final velocity \( v \), acceleration \( a \), and distance \( s \) is \( v^2 = u^2 + 2as \). We know the mower comes to a stop, so \( v = 0 \, \text{m/s} \), the initial velocity \( u = 1.5 \, \text{m/s} \), and the acceleration \( a = -1 \, \text{m/s}^2 \). We need to solve for \( s \).

Rearranging the kinematic equation for \( s \): \( s = \frac{v^2 - u^2}{2a} \)

Step 3: Substitute the known values into the kinematic equation

Substituting \( v = 0 \), \( u = 1.5 \, \text{m/s} \), and \( a = -1 \, \text{m/s}^2 \) into the formula:

\( s = \frac{0 - (1.5)^2}{2(-1)} = \frac{-2.25}{-2} = 1.125 \, \text{m} \)

Final Answers

• The force \( F \) exerted by the person is \( \boldsymbol{75 \, \text{N}} \).
• The distance the mower travels before stopping is \( \boldsymbol{1.125 \, \text{m}} \) (or \( 1.13 \, \text{m} \) if rounded to two decimal places).