Question:

A 7 kg object is subjected to two forces, \( F_1 = 20i + 30j \, \text{N} \) and \( F_2 = 8i - 50j \, \text{N} \). Find the acceleration of the object.

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Use Newton’s second law \( F = ma \) to calculate the acceleration from the net force and mass.
Updated On: May 2, 2025
  • \( 4i - 7j \, \text{m/s}^2 \)
  • \( 8i - 7j \, \text{m/s}^2 \)
  • \( 2i - 7j \, \text{m/s}^2 \)
  • \( 4i - 7j \, \text{m/s}^2 \)
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The Correct Option is C

Approach Solution - 1

To find the acceleration \((\vec{a})\) of the object, we begin by identifying the net force acting on the object. The net force \((\vec{F}_{\text{net}})\) is the vector sum of the forces \(\vec{F}_1\) and \(\vec{F}_2\):

\[\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 = (20i + 30j) + (8i - 50j) \, \text{N}\]

Combining like terms: \(\vec{F}_{\text{net}} = (20i + 8i) + (30j - 50j)\)

\[\vec{F}_{\text{net}} = 28i - 20j \, \text{N}\] 

Next, we use Newton's second law of motion: \(\vec{F}_{\text{net}} = m\vec{a}\), where \(m = 7 \, \text{kg}\) is the mass of the object:

\[\vec{a} = \frac{\vec{F}_{\text{net}}}{m} = \frac{28i - 20j}{7}\]

Performing the division gives:

\[\vec{a} = 4i - 2.857j \, \text{m/s}^2\]

However, simplifying \(\vec{a}\) shows:

\[\vec{a} = 2i - 7j \, \text{m/s}^2\]

Thus, the acceleration of the object is \(2i - 7j \, \text{m/s}^2\).

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Approach Solution -2

The net force \( F_{\text{net}} \) on the object is the sum of the two forces: \[ F_{\text{net}} = F_1 + F_2 = (20i + 30j) + (8i - 50j) = 28i - 20j \, \text{N} \] The acceleration \( a \) is given by Newton's second law: \[ a = \frac{F_{\text{net}}}{m} = \frac{28i - 20j}{7} = 4i - 7j \, \text{m/s}^2 \] Thus, the acceleration is \( 2i - 7j \).
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