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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 single conservative force \(\mathrm{F}(\mathrm{x})\) acts on a $2.5 \mathrm{~kg}\( particle that moves along the \)\mathrm{x}$ -axis. The potential energy \(\mathrm{U}(\mathrm{x})\) is given by \(\mathrm{U}(\mathrm{x})=\left[10+(\mathrm{x}-4)^{2}\right]\) where \(\mathrm{x}\) is in meter. \(\mathrm{At} \mathrm{x}=6.0 \mathrm{~m}\) the particle has kinetic energy of \(20 \mathrm{~J}\). what is the mechanical energy of the system? (A) \(34 \mathrm{~J}\) (B) \(45 \mathrm{~J}\) (C) \(48 \mathrm{~J}\) (D) \(49 \mathrm{~J}\)

A bomb of mass \(3.0 \mathrm{~kg}\) explodes in air into two pieces of masses \(2.0 \mathrm{~kg}\) and \(1.0 \mathrm{~kg}\). The smaller mass goes at a speed of \(80 \mathrm{~m} / \mathrm{s}\). The total energy imparted to the two fragments is (A) \(1.07 \mathrm{KJ}\) (B) \(2.14 \mathrm{KJ}\) (C) \(2.4 \mathrm{KJ}\) (D) \(4.8 \mathrm{KJ}\)

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If both assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are false. Assertion: A weight lifter does no work in holding the weight up. Reason: Work done is zero because distance moved is zero. (A) 1 (B) 2 (C) 3 (D) 4

A metal ball of mass \(2 \mathrm{~kg}\) moving with a velocity of $36 \mathrm{~km} / \mathrm{h}$ has a head on collision with a stationary ball of mass \(3 \mathrm{~kg}\). If after the collision, the two balls move together, the loss in kinetic energy due to collision is (A) \(40 \mathrm{~J}\) (B) \(60 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(140 \mathrm{~J}\)

An engine pump is used to pump a liquid of density \(\rho\) continuously through a pipe of cross-sectional area \(\mathrm{A}\). If the speed of flow of the liquid in the pipe is \(\mathrm{v}\), then the rate at which kinetic energy is being imparted to the liquid is (A) \((1 / 2) \mathrm{A} \rho \mathrm{V}^{3}\) (B) \((1 / 2) \mathrm{A} \rho \mathrm{V}^{2}\) (C) \((1 / 2) \mathrm{A} \rho \mathrm{V}\) (B) \(\mathrm{ApV}\)
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