Chapter 4: Problem 420
A force of \(7 \mathrm{~N}\), making an angle \(\theta\) with the horizontal, acting on an object displaces it by \(0.5 \mathrm{~m}\) along the horizontal direction. If the object gains K.E. of \(2 \mathrm{~J}\), what is the horizontal component of the force? (A) \(2 \mathrm{~N}\) (B) \(4 \mathrm{~N}\) (C) \(1 \mathrm{~N}\) (D) \(14 \mathrm{~N}\)
Short Answer
Step by step solution
Recall the work-energy theorem
Calculate the work done
Recall the formula for work done
Substitute the known values
Calculate the horizontal component of the force
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinetic Energy
- \( KE \) is the kinetic energy.
- \( m \) represents the mass of the object in kilograms.
- \( v \) is the velocity of the object in meters per second.
Components of Force
- The horizontal component: \( F_x = F \cos{\theta} \)
- The vertical component: \( F_y = F \sin{\theta} \)
Calculating Work Done
- \( W \) is the work done in joules.
- \( F \) is the magnitude of the force in newtons.
- \( d \) is the displacement in meters.
- \( \theta \) is the angle between the force and the displacement direction.