Question:
A force \(\vec{F_1}\) of magnitude \(4\;N\) acts on an object of mass \(1\;kg\), at origin in a direction \(30^\circ\) above the positive \(x\)-axis. A second force \(\vec{F_2}\) of magnitude \(4\;N\) acts on the same object in the direction of the positive \(y\)-axis. The magnitude of the acceleration of the object is nearly
A force \(\vec{F_1}\) of magnitude \(4\;N\) acts on an object of mass \(1\;kg\), at origin in a direction \(30^\circ\) above the positive \(x\)-axis. A second force \(\vec{F_2}\) of magnitude \(4\;N\) acts on the same object in the direction of the positive \(y\)-axis. The magnitude of the acceleration of the object is nearly
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When forces act at angles, first resolve them into horizontal and vertical components, then find the resultant using the Pythagoras theorem.
Updated On: Jun 22, 2026
- \(6.9\;\text{m s}^{-2}\)
- \(7.6\;\text{m s}^{-2}\)
- \(4.3\;\text{m s}^{-2}\)
- \(8.0\;\text{m s}^{-2}\)
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The Correct Option is A
Solution and Explanation
Step 1: Resolve the first force into components.
The force \(\vec{F_1}\) has magnitude \(4\;N\) and acts at an angle \(30^\circ\) above the positive \(x\)-axis.
Its \(x\)-component is \[ F_{1x}=4\cos30^\circ \] \[ =4\left(\frac{\sqrt{3}}{2}\right) \] \[ =2\sqrt{3}\;N \] Its \(y\)-component is \[ F_{1y}=4\sin30^\circ \] \[ =4\left(\frac12\right) \] \[ =2\;N \]
Step 2: Write the components of the second force.
The second force \(\vec{F_2}\) acts along the positive \(y\)-axis with magnitude \(4\;N\).
Therefore, \[ F_{2x}=0 \] and \[ F_{2y}=4\;N \]
Step 3: Find the resultant force.
Total \(x\)-component of force: \[ F_x=2\sqrt{3} \] Total \(y\)-component of force: \[ F_y=2+4 \] \[ =6 \] Magnitude of resultant force: \[ F=\sqrt{F_x^2+F_y^2} \] \[ =\sqrt{(2\sqrt{3})^2+6^2} \] \[ =\sqrt{12+36} \] \[ =\sqrt{48} \] \[ =4\sqrt{3}\;N \] \[ \approx 6.93\;N \]
Step 4: Use Newton's second law.
Given mass, \[ m=1\;kg \] Acceleration is \[ a=\frac{F}{m} \] \[ =\frac{4\sqrt{3}}{1} \] \[ =4\sqrt{3}\;\text{m s}^{-2} \] \[ \approx 6.9\;\text{m s}^{-2} \]
Step 5: Final conclusion.
Hence, the magnitude of acceleration is \[ \boxed{6.9\;\text{m s}^{-2}} \]
The force \(\vec{F_1}\) has magnitude \(4\;N\) and acts at an angle \(30^\circ\) above the positive \(x\)-axis.
Its \(x\)-component is \[ F_{1x}=4\cos30^\circ \] \[ =4\left(\frac{\sqrt{3}}{2}\right) \] \[ =2\sqrt{3}\;N \] Its \(y\)-component is \[ F_{1y}=4\sin30^\circ \] \[ =4\left(\frac12\right) \] \[ =2\;N \]
Step 2: Write the components of the second force.
The second force \(\vec{F_2}\) acts along the positive \(y\)-axis with magnitude \(4\;N\).
Therefore, \[ F_{2x}=0 \] and \[ F_{2y}=4\;N \]
Step 3: Find the resultant force.
Total \(x\)-component of force: \[ F_x=2\sqrt{3} \] Total \(y\)-component of force: \[ F_y=2+4 \] \[ =6 \] Magnitude of resultant force: \[ F=\sqrt{F_x^2+F_y^2} \] \[ =\sqrt{(2\sqrt{3})^2+6^2} \] \[ =\sqrt{12+36} \] \[ =\sqrt{48} \] \[ =4\sqrt{3}\;N \] \[ \approx 6.93\;N \]
Step 4: Use Newton's second law.
Given mass, \[ m=1\;kg \] Acceleration is \[ a=\frac{F}{m} \] \[ =\frac{4\sqrt{3}}{1} \] \[ =4\sqrt{3}\;\text{m s}^{-2} \] \[ \approx 6.9\;\text{m s}^{-2} \]
Step 5: Final conclusion.
Hence, the magnitude of acceleration is \[ \boxed{6.9\;\text{m s}^{-2}} \]
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