An object is acted upon by the forces `vec(F)_(1)=4hat(i)N` and `vec(F)_(2)=(hat(i)-hat(j))N`. If the displacement of the object is `D=(hat(i)+6hat(j)-hat(k))`m, the kinetic energy of the object

Options

A. remains constant
B. increases by 1 J
C. decreases by 1 J
D. decreases by 2 J

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the Forces
We have two forces acting on the object:
- \( \vec{F}_1 = 4 \hat{i} \, \text{N} \)
- \( \vec{F}_2 = \hat{i} - \hat{j} \, \text{N} \)

### Step 2: Calculate the Net Force
To find the net force \( \vec{F} \), we add the two forces:
\[ \vec{F} = \vec{F}_1 + \vec{F}_2 = (4 \hat{i}) + (\hat{i} - \hat{j}) = (4 + 1) \hat{i} - \hat{j} = 5 \hat{i} - \hat{j} \, \text{N} \]

### Step 3: Identify the Displacement
The displacement vector is given as:
\[ \vec{D} = \hat{i} + 6 \hat{j} - \hat{k} \, \text{m} \]

### Step 4: Calculate the Work Done
The work done \( W \) by the net force on the object is given by the dot product of the net force and the displacement:
\[ W = \vec{F} \cdot \vec{D} = (5 \hat{i} - \hat{j}) \cdot (\hat{i} + 6 \hat{j} - \hat{k}) \]
Calculating the dot product:
\[ W = 5 \cdot 1 + (-1) \cdot 6 + 0 \cdot (-1) = 5 - 6 + 0 = -1 \, \text{J} \]

### Step 5: Relate Work to Change in Kinetic Energy
According to the work-energy theorem, the work done is equal to the change in kinetic energy (\( \Delta KE \)):
\[ \Delta KE = W = -1 \, \text{J} \]

### Step 6: Conclusion
The kinetic energy of the object has decreased by 1 Joule.

### Final Answer
The change in kinetic energy is \(-1 \, \text{J}\).

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