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Chapter 4: Problem 484

The decreases in the potential energy of a ball of mass \(25 \mathrm{~kg}\) which falls from a height of \(40 \mathrm{~cm}\) is (A) \(968 \mathrm{~J}\) (B) \(100 \mathrm{~J}\) (C) \(1980 \mathrm{~J}\) (D) \(200 \mathrm{~J}\)

Short Answer

Expert verified
The decrease in potential energy is calculated using the formula \(PE = m * g * h\). Given the mass (m) of the ball as \(25 kg\) and the height (h) as \(40 cm\), we first convert the height to meters (\(0.4 m\)). The initial potential energy is \(PE_{initial} = (25 kg) * (9.8 \frac{m}{s^2}) * (0.4 m) = 98 Nm\). Since the ball falls to the ground, the final potential energy is \(0 Nm\). The decrease in potential energy is \(PE_{initial} - PE_{final} = 98 Nm - 0 Nm = 98 Nm\). None of the given options match this value, so the correct answer should be approximately \(98 J\).

Step by step solution

01

Understand the problem and recall the formula for potential energy

We are given the mass (m) of the ball as 25 kg and the height (h) it falls from as 40 cm. We need to find the decrease in potential energy, which can be calculated using the formula for potential energy: Potential energy (PE) = m * g * h, where g represents the acceleration due to gravity (approximately 9.8 m/s²).
02

Convert height to meters

We are given the height in centimeters, so we need to convert it to meters before performing the calculation: \(h = 40 cm = 0.4 m\)
03

Calculate the initial potential energy

Now we can calculate the initial potential energy of the ball before it falls, using the formula PE = m * g * h with the provided values for mass and height, and a value of 9.8 m/s² for acceleration due to gravity: \(PE_{initial} = (25 kg) * (9.8 \frac{m}{s^2}) * (0.4 m) = 98 Nm\)
04

Calculate the final potential energy

Since the ball falls to the ground, its final height is 0 meters. Therefore, the final potential energy will be zero: \(PE_{final} = (25 kg) * (9.8 \frac{m}{s^2}) * (0 m) = 0 Nm\)
05

Determine the decrease in potential energy

Finally, to find the decrease in potential energy, we subtract the final potential energy from the initial potential energy: \(Decrease~in~potential~energy = PE_{initial} - PE_{final} = 98 Nm - 0 Nm = 98 Nm\) Comparing this value with the given options: (A) \(968 \mathrm{~J}\) (B) \(100 \mathrm{~J}\) (C) \(1980 \mathrm{~J}\) (D) \(200 \mathrm{~J}\) None of these options match the decrease in potential energy calculated. There may be an error in the given options, or the problem statement might be incorrect. Based on the calculation, the correct answer should be approximately 98 J.

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Most popular questions from this chapter

The velocity of a body of mass \(400 \mathrm{gm}\) is $(-3 \mathrm{i} \wedge-4 \mathrm{j} \wedge) \mathrm{m} / \mathrm{s}$. So its kinetic energy is ...... (A) \(5 \mathrm{~J}\) (B) \(10 \mathrm{~J}\) (C) \(8 \mathrm{~J}\) (D) \(16 \mathrm{~J}\)

Four identical balls are lined in a straight grove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide elastically with the row of 4 balls from left. What will happen (A) One ball from the right rolls out with a speed \(2 \mathrm{v}\) and the remaining balls will remain at rest. (B) Two balls from the right roll out speed \(\mathrm{v}\) each and the remaining balls will remain stationary. (C) All the four balls in the row will roll out with speed \(\mathrm{v}(\mathrm{v} / 4)\) each and the two colliding balls will come to rest. (D) The colliding balls will come to rest and no ball rolls out from right.

A body of mass \(\mathrm{m}\) is accelerated uniformly from rest to a speed \(\mathrm{v}\) in time \(\mathrm{T}\). The instantaneous power delivered to the body in terms of time is given by..... (A) $\left[\left\\{\mathrm{mv}^{2}\right\\} /\left\\{\mathrm{T}^{2}\right\\}\right] \cdot \mathrm{t}$ (B) $\left[\left\\{\mathrm{mv}^{2}\right\\} /\left\\{\mathrm{T}^{2}\right\\}\right] \cdot \mathrm{t}^{2}$ (C) $\left[\left\\{\mathrm{mv}^{2}\right\\} /\\{2 \mathrm{~T}\\}\right] \cdot \mathrm{t}$ (D) $\left[\left\\{\mathrm{mv}^{2}\right\\} /\left\\{2 \mathrm{~T}^{2}\right\\}\right] \cdot \mathrm{t}^{2}$

A force \(F=A y^{2}+B y+C\) acts on a body in the \(y\) -direction. The work done by this force during a displacement from \(\mathrm{y}=-\mathrm{a}\) to \(\mathrm{y}=\mathrm{a}\) is (A) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]\) (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\) (C) $\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+\left[\left\\{\mathrm{Ba}^{2}\right\\} / 2\right]+\mathrm{ca}$ (D) None of these.

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) \(140 \mathrm{~J}\)
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