A block is kept on x-y plane at a point (2m,3m). It is pushed with a force of magnitude 10 Ntowards point (5m,7m). The force can be represented as?

To determine how the force applied to the block can be represented, we first need to analyze the situation in terms of vectors. The block is initially located at the point (2m, 3m) on the x-y plane, and it is being pushed towards the point (5m, 7m) with a force of 10 N. Let's break this down step by step.

Understanding the Force Vector

A force can be represented as a vector, which has both magnitude and direction. In this case, the magnitude of the force is given as 10 N. To find the direction, we need to calculate the vector that points from the initial position of the block to the target position.

Calculating the Direction Vector

The direction vector can be found by subtracting the coordinates of the initial position from the coordinates of the target position:

• Initial position: (2, 3)
• Target position: (5, 7)

The components of the direction vector (Δx, Δy) are calculated as follows:

• Δx = 5 - 2 = 3
• Δy = 7 - 3 = 4

Thus, the direction vector is (3, 4).

Normalizing the Direction Vector

To represent the force vector accurately, we need to normalize the direction vector. Normalization involves converting the vector into a unit vector, which has a magnitude of 1. The magnitude of the direction vector can be calculated using the Pythagorean theorem:

Magnitude = √(Δx² + Δy²) = √(3² + 4²) = √(9 + 16) = √25 = 5

Now, we can find the unit vector (u) by dividing each component of the direction vector by its magnitude:

• u_x = Δx / Magnitude = 3 / 5 = 0.6
• u_y = Δy / Magnitude = 4 / 5 = 0.8

So, the unit vector is (0.6, 0.8).

Calculating the Force Vector

Now that we have the unit vector, we can calculate the force vector (F) by multiplying the unit vector by the magnitude of the force:

• F_x = 10 N * 0.6 = 6 N
• F_y = 10 N * 0.8 = 8 N

Therefore, the force vector can be represented as:

F = (6 N, 8 N)

Visualizing the Force

To visualize this, imagine the block at (2, 3) on a coordinate plane. The force vector (6 N, 8 N) points in the direction of (5, 7), indicating that the block is being pushed diagonally upwards to the right. The force effectively moves the block towards the target point while overcoming any friction or resistance it may encounter.

Summary

In summary, the force applied to the block can be represented as a vector with components (6 N, 8 N). This representation not only indicates the magnitude of the force but also its direction, allowing for a comprehensive understanding of how the block will move in response to the applied force.