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Chapter 4: Problem 451

A body of mass \(1 \mathrm{~kg}\) is thrown upwards with a velocity $20 \mathrm{~m} / \mathrm{s}\(. It momentarily comes to rest after a height \)18 \mathrm{~m}\(. How much energy is lost due to air friction. \)(\mathrm{g}=10 \mathrm{~m} / \mathrm{s} 2)$ (A) \(20 \mathrm{~J}\) (B) \(30 \mathrm{~J}\) (C) \(40 \mathrm{~J}\) (D) \(10 \mathrm{~J}\)

Short Answer

Expert verified
The energy lost due to air friction is \(20 \mathrm{~J}\), which corresponds to option (A).

Step by step solution

01

Calculate initial kinetic energy (KE)

We can calculate the initial kinetic energy of the body using the formula KE = \(\frac{1}{2}\)mv², where m is the mass of the body (1 kg) and v is its initial velocity (20 m/s). This will give us the initial kinetic energy in Joules (J). KE = \(\frac{1}{2}(1 \mathrm{~kg})(20 \mathrm{~m/s})^2\) = \(200 \mathrm{~J}\)
02

Calculate final potential energy (PE)

Now, we need to find the final potential energy of the body when it reaches the height of 18 m. We can do this using the formula PE = mgh, where m is the mass (1 kg), g is the gravitational acceleration (10 m/s²), and h is the height (18 m). PE = (1 \mathrm{~kg})(10 \mathrm{~m/s²})(18 \mathrm{~m}) = \(180 \mathrm{~J}\)
03

Calculate net energy at the highest point

When the body momentarily comes to rest, its final kinetic energy will be 0 J, as it has stopped completely. Therefore, the net energy at the highest point will be equal to the final potential energy. Net energy = PE = \(180 \mathrm{~J}\)
04

Determine the energy lost due to air friction

To find the energy lost due to air friction, we need to subtract the net energy at the highest point from the initial kinetic energy. Energy loss = Initial KE - Net Energy = \(200 \mathrm{~J} - 180 \mathrm{~J} = 20 \mathrm{~J}\) So, the energy lost due to air friction is \(20 \mathrm{~J}\), which corresponds to option (A).

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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) \(10 \sqrt{12} \mathrm{~m} / \mathrm{s}\)

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

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: Both, a stretched spring and a compressed spring have potential energy. Reason: Work is done against the restoring force in each case. (A) 1 (B) 2 (C) 3 (D) 4

How much is the work done in pulling up a block of wood weighing $2 \mathrm{KN}\( for a length of \)10 \mathrm{~m}$ on a smooth plane inclined at an angle of \(30^{\circ}\) with the horizontal? (A) \(1.732 \mathrm{KJ}\) (B) \(17.32 \mathrm{KJ}\) (C) \(10 \mathrm{KJ}\) (D) \(100 \mathrm{KJ}\)

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) \(4.8 \mathrm{KJ}\)
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