An object of mass 5 kg is attached to the end ofa rope. If the rope is pulled upward with an acceleration `0.30 ms^(-2),` what is the tension in the rope?

Options

A. 30.5 N
B. 40.5 N
C. 50.5 N
D. 60.5 N

Explanation

To find the tension in the rope when an object of mass 5 kg is pulled upward with an acceleration of 0.30 m/s², we can follow these steps:

### Step 1: Identify the forces acting on the object
The forces acting on the object are:
- The weight of the object (downward force), which is given by the formula:
\[ W = mg \]
where \( m = 5 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).

- The tension in the rope (upward force), denoted as \( T \).

### Step 2: Calculate the weight of the object
Using the formula for weight:
\[ W = mg = 5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 49 \, \text{N} \]

### Step 3: Apply Newton's second law of motion
According to Newton's second law, the net force acting on the object is equal to the mass of the object multiplied by its acceleration:
\[ F_{\text{net}} = ma \]
In this case, the net force can be expressed as:
\[ T - W = ma \]
Substituting the known values:
\[ T - 49 \, \text{N} = 5 \, \text{kg} \times 0.30 \, \text{m/s}^2 \]

### Step 4: Calculate the net force
Calculating the right side:
\[ 5 \, \text{kg} \times 0.30 \, \text{m/s}^2 = 1.5 \, \text{N} \]
So we have:
\[ T - 49 \, \text{N} = 1.5 \, \text{N} \]

### Step 5: Solve for tension \( T \)
Rearranging the equation to solve for \( T \):
\[ T = 49 \, \text{N} + 1.5 \, \text{N} = 50.5 \, \text{N} \]

### Final Answer
The tension in the rope is \( T = 50.5 \, \text{N} \).

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