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Chapter 3: Problem 348

A \(7 \mathrm{~kg}\) object is subjected to two forces (in newton) \(\underline{F}_{1}=20 \mathrm{i}^{-}+30 \mathrm{j}^{-}\) and \(\underline{\mathrm{F}}_{2}=8 \mathrm{i}^{-}-5 \mathrm{j}\) `The magnitude of resulting acceleration in \(\mathrm{ms}^{-2}\) will be (A) 5 (B) 4 (C) 3 (D) 2

Short Answer

Expert verified
The magnitude of the resulting acceleration is approximately 5 \(\mathrm{ms}^{-2}\).

Step by step solution

01

Find the net force acting on the object.

To find the net force acting on the object, we need to add the two given forces: \(\underline{F}_{1}=20 \mathrm{i}^{-}+30 \mathrm{j}^{-}\) and \(\underline{\mathrm{F}}_{2}=8 \mathrm{i}^{-}-5 \mathrm{j}\) \[ \underline{F}_{net} = \underline{F}_{1} + \underline{F}_{2} \]
02

Add the forces component-wise.

To add the forces, just add their corresponding components: \[ \underline{F}_{net} = [(20 + 8) \mathrm{i}^{-} +(30 - 5) \mathrm{j}^{-}] \] \[ \underline{F}_{net} = [28\mathrm{~i}^{-}+ 25 \mathrm{~j}^{-}] \]
03

Apply Newton's second law of motion.

According to Newton's second law, the force acting on an object is equal to the mass of the object times its acceleration: \[ \underline{F}_{net} = m \cdot \underline{a} \] Using the given mass of the object (m=7kg) and the calculated net force, we can now determine the acceleration: \[ 28\mathrm{~i}^{-}+ 25 \mathrm{~j}^{-} = 7 \mathrm{~kg} \cdot \underline{a} \]
04

Find the acceleration vector.

To find the acceleration vector, divide the net force vector by the mass: \[ \underline{a} = \frac{\underline{F}_{net}}{m} = \frac{28\mathrm{~i}^{-} + 25\mathrm{~j}^{-}}{7\mathrm{~kg}} \] \[ \underline{a} = 4 \mathrm{i}^{-} + \frac{25}{7} \mathrm{j}^{-} \]
05

Calculate the magnitude of the acceleration vector.

Now we can find the magnitude of the acceleration vector using the Pythagorean theorem: \[ | \underline{a} | = \sqrt{(4)^2 + (\frac{25}{7})^2} \] Calculate the magnitude: \[ | \underline{a} | = \sqrt{16 + (\frac{625}{49})} \] \[ | \underline{a} | = \sqrt{\frac{1577}{49}} \approx 5.7 \mathrm{m/s}^{2} \] From the four options given, the closest answer is (A) 5. So, the magnitude of the resulting acceleration is approximately 5 \(\mathrm{ms}^{-2}\).

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