Problem 420 - JEE Physics

Question

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) \(14 \mathrm{~N}\)

Step-by-Step Solution

Answer

(B) \(4 \mathrm{~N}\)

Step 1: Recall the work-energy theorem

The work-energy theorem states that the work done on an object is equal to the change in the object's kinetic energy: \[W = \Delta KE\]

Step 2: Calculate the work done

Since the object gains a kinetic energy of 2 J, the work done by the force on the object is also 2 J: \[W = 2 \mathrm{~J}\]

Step 3: Recall the formula for work done

The work done by a force on an object is given by: \[W = Fd \cos{\theta}\] Where \(F\) is the magnitude of the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and displacement.

Step 4: Substitute the known values

We know that the total force has a magnitude of 7 N, the displacement is 0.5 m, and the work done is 2 J. Let's substitute these values into the formula: \[2 \mathrm{~J} = (7 \mathrm{~N})(0.5 \mathrm{~m}) \cos{\theta}\]

Step 5: Calculate the horizontal component of the force

The horizontal component of the force is given by: \[F_x = F \cos{\theta}\] Since we want to find the horizontal component of the force, we need to calculate \(F_x\) from the equation above. First, let's rewrite the equation to find \(\cos{\theta}\): \[\cos{\theta} = \frac{2 \mathrm{~J}}{(7 \mathrm{~N})(0.5 \mathrm{~m})}\] Now, substitute this into the formula for the horizontal component of the force: \[F_x = (7 \mathrm{~N})\left(\frac{2 \mathrm{~J}}{(7 \mathrm{~N})(0.5 \mathrm{~m})}\right)\] Now simplify this expression: \[F_x = 4 \mathrm{~N}\] So the horizontal component of the force is 4 N. The correct answer is (B) \(4 \mathrm{~N}\).

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. It is directly proportional to the mass of the object and the square of its velocity. The formula for kinetic energy is given by:\[ KE = \frac{1}{2}mv^2 \]where:

• \( KE \) is the kinetic energy.
• \( m \) represents the mass of the object in kilograms.
• \( v \) is the velocity of the object in meters per second.

Kinetic energy is a scalar quantity, meaning it has magnitude but no direction. In the context of the work-energy theorem, if an object gains kinetic energy, work has been done on it. This change in kinetic energy indicates the energy transferred to the object by a force across a distance.

Components of Force

Forces can act in many directions, but often it's useful to break them down into components along the coordinate axes, usually horizontal and vertical. This is done using trigonometry when the force is applied at an angle.The components of a force \( F \) at an angle \( \theta \) to the horizontal are:

• The horizontal component: \( F_x = F \cos{\theta} \)
• The vertical component: \( F_y = F \sin{\theta} \)

In problems like this, we often focus on the component of force that acts in the direction of movement since it contributes to the work done on the object. By decomposing the force into its components, we can analyze its effects more clearly. The horizontal component, in particular, plays a key role in determining the work done on the object in the direction of displacement.

Calculating Work Done

Work is done when a force moves an object through a distance. It can be calculated using the formula:\[ W = Fd \cos{\theta} \]where:

• \( W \) 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.

The cosine function captures the component of the force that is effective in doing work in the displacement's direction. When \(\theta\) is 0 degrees, the full magnitude of the force contributes to work. If the angle is greater, less of the force contributes directly to the movement, reducing the work done. When we apply this formula correctly, we determine the amount of energy transferred, as seen in changes in the object's kinetic energy.

Displacement

Displacement refers to the change in position of an object. It is a vector quantity, implying it has both magnitude and direction. Displacement is different from distance in that it considers the initial and final positions, not the path taken. For directional work calculations, only the displacement in the direction of the force matters. In the problem we considered, the object was displaced by 0.5 meters along the horizontal plane. This means the calculated work is based on this horizontal movement, relating directly to the force component in the same direction. Understanding displacement helps create clarity in problems involving the work-energy principle, as it essentially connects how far the object has moved as a result of the force applied.