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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

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

Step by step solution

01

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: \[W = \Delta KE\]
02

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: \[W = 2 \mathrm{~J}\]
03

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.
04

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: \[2 \mathrm{~J} = (7 \mathrm{~N})(0.5 \mathrm{~m}) \cos{\theta}\]
05

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

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Most popular questions from this chapter

A spherical ball of mass \(15 \mathrm{~kg}\) stationary at the top of a hill of height \(82 \mathrm{~m} .\) It slides down a smooth surface to the ground, then climbs up another hill of height \(32 \mathrm{~m}\) and finally slides down to horizontal base at a height of \(10 \mathrm{~m}\) above the ground. The velocity attained by the ball is (A) \(30 \sqrt{10 \mathrm{~m} / \mathrm{s}}\) (B) \(10 \sqrt{30 \mathrm{~m} / \mathrm{s}}\) (C) \(12 \sqrt{10} \mathrm{~m} / \mathrm{s}\) (D) \(10 \sqrt{12} \mathrm{~m} / \mathrm{s}\)

A ball is allowed to fall from a height \(20 \mathrm{~m}\). If there is \(30 \%\) loss of energy due to impact, then after one impact ball will go up to (A) \(18 \mathrm{~m}\) (B) \(16 \mathrm{~m}\) (C) \(12 \mathrm{~m}\) (D) \(14 \mathrm{~m}\)

A spring gun of spring constant \(90 \times 10^{2} \mathrm{~N} / \mathrm{M}\) is compressed \(4 \mathrm{~cm}\) by a ball of mass \(16 \mathrm{~g}\). If the trigger is pulled, calculate the velocity of the ball. (A) \(60 \mathrm{~m} / \mathrm{s}\) (B) \(3 \mathrm{~m} / \mathrm{s}\) (C) \(90 \mathrm{~m} / \mathrm{s}\) (D) \(30 \mathrm{~m} / \mathrm{s}\)

The potential energy of \(2 \mathrm{~kg}\) particle, free to move along \(\mathrm{x}\) axis is given by $\mathrm{U}(\mathrm{X})=\left[\left\\{\mathrm{x}^{4} / 4\right\\}-\left\\{\mathrm{x}^{2} / 2\right\\}\right] \mathrm{J}$. If its mechanical energy is \(2 \mathrm{~J}\), its maximum speed is $\ldots \mathrm{m} / \mathrm{s}$ (A) \((3 / 2)\) (B) \(\sqrt{2}\) (C) \((1 / \sqrt{2})\) (D) 2

If the K.E. of a body is increased by \(44 \%\), its momentum will increase by....... (A) \(20 \%\) (B) \(22 \%\) (C) \(2 \%\) (D) \(120 \%\)
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