Answer :
To solve this problem, we are trying to find the magnitude of the pulling force [tex]\( P \)[/tex] such that the net work done on the crate is zero. This means that the horizontal component of the force [tex]\( P \)[/tex] should exactly balance the frictional force. Here's how you can approach the problem step-by-step:
1. Understand the Forces Involved:
- The crate is being pulled by a force [tex]\( P \)[/tex] that makes a 30° angle with the horizontal floor.
- There is a frictional force of 196 N opposing the motion of the crate.
2. Work Done on the Crate:
- For the net work done on the crate to be zero, the horizontal component of the force [tex]\( P \)[/tex] must be equal to the frictional force. This is because work is done when there is a net force causing displacement. If the net force is zero (forces balance), there is no net work.
3. Resolve the Force [tex]\( P \)[/tex] into Components:
- The horizontal component of [tex]\( P \)[/tex] is given by [tex]\( P \cos(\theta) \)[/tex], where [tex]\( \theta = 30^\circ \)[/tex].
4. Set the Horizontal Component Equal to the Frictional Force:
- To achieve zero net work, the horizontal component of the pulling force must equal the frictional force.
[tex]\[
P \cos(30^\circ) = 196 \, \text{N}
\][/tex]
5. Solve for [tex]\( P \)[/tex]:
- Using the cosine of 30° which is [tex]\(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)[/tex], solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{196}{\cos(30^\circ)}
\][/tex]
6. Calculate the Magnitude of [tex]\( P \)[/tex]:
- Substituting the cosine value:
[tex]\[
P = \frac{196}{\frac{\sqrt{3}}{2}}
\][/tex]
This results in:
[tex]\[
P \approx 226.32 \, \text{N}
\][/tex]
Thus, the magnitude of the force [tex]\( P \)[/tex] needed to ensure the net work done on the crate is zero is approximately 226.32 N. This means the horizontal component of this force perfectly balances the frictional force on the crate.
1. Understand the Forces Involved:
- The crate is being pulled by a force [tex]\( P \)[/tex] that makes a 30° angle with the horizontal floor.
- There is a frictional force of 196 N opposing the motion of the crate.
2. Work Done on the Crate:
- For the net work done on the crate to be zero, the horizontal component of the force [tex]\( P \)[/tex] must be equal to the frictional force. This is because work is done when there is a net force causing displacement. If the net force is zero (forces balance), there is no net work.
3. Resolve the Force [tex]\( P \)[/tex] into Components:
- The horizontal component of [tex]\( P \)[/tex] is given by [tex]\( P \cos(\theta) \)[/tex], where [tex]\( \theta = 30^\circ \)[/tex].
4. Set the Horizontal Component Equal to the Frictional Force:
- To achieve zero net work, the horizontal component of the pulling force must equal the frictional force.
[tex]\[
P \cos(30^\circ) = 196 \, \text{N}
\][/tex]
5. Solve for [tex]\( P \)[/tex]:
- Using the cosine of 30° which is [tex]\(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)[/tex], solve for [tex]\( P \)[/tex]:
[tex]\[
P = \frac{196}{\cos(30^\circ)}
\][/tex]
6. Calculate the Magnitude of [tex]\( P \)[/tex]:
- Substituting the cosine value:
[tex]\[
P = \frac{196}{\frac{\sqrt{3}}{2}}
\][/tex]
This results in:
[tex]\[
P \approx 226.32 \, \text{N}
\][/tex]
Thus, the magnitude of the force [tex]\( P \)[/tex] needed to ensure the net work done on the crate is zero is approximately 226.32 N. This means the horizontal component of this force perfectly balances the frictional force on the crate.
