Question

A 4 kg object is given a displacement Δ s = (4 m)ˆı (2 m) ˆ (-5 m) ˆk along a straight line. During the displacement, a constant force of f = (2 n)ˆı (-3 n) ˆ (1 n) ˆk acts on the object. Find the work done by f for this displacement. Answer in units of J.

Answer 1

The work done by the force vector f on the object with the displacement Δs is -3 J, showing that the force acts opposite the displacement.

Step-by-step explanation:

To calculate the work done by a force during a displacement, you use the dot product of the force vector and the displacement vector. The dot product is calculated by multiplying the corresponding components of the vectors and summing the results. In formula form, Work (W) is W = F · Δs, where F is the force and Δs is the displacement.

In this case, the force vector is f = (2 N)´® + (-3 N)¯¯º + (1 N)®º, and the displacement vector is Δs = (4 m)´® + (2 m)¯¯º + (-5 m)®º. Calculating the dot product, we find:

W = (2 N × 4 m) + (-3 N × 2 m) + (1 N × -5 m)
W = 8 J - 6 J - 5 J
W = -3 J

The negative sign indicates that on average the direction of the force is opposite to the direction of the displacement.