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Chapter 6: Problem 2

A ball of mass \(0.2 \mathrm{~kg}\) is thrown vertically upwards by applying a force by hand. If the hand moves \(0.2 \mathrm{~m}\) while applying a force and the ball goes upto \(2 \mathrm{~m}\) height further find, the magnitude of the force. Consider \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) [a) \(16 \mathrm{~N}\) (b) \(20 \mathrm{~N}\) (c) \(22 \mathrm{~N}\) (d) \(44 \mathrm{~N}\)

Short Answer

Expert verified
The magnitude of the force is 22 N (option c).

Step by step solution

01

Understand the problem

A ball is thrown upwards with an initial force, moves a certain distance while in contact with the force, then continues moving upwards without the force. We need to determine the force applied by the hand to achieve the ball's upward motion.
02

Apply Work-Energy Principle

The work done by the hand when it applies a force is equal to the change in potential energy of the ball. Work can be expressed as \[ W = F \times d \] where \( F \) is the force and \( d \) is the distance over which the force is applied. The potential energy change due to the ball's ascent (from the base to maximum height, including both segments) is given by \[ \Delta U = m \times g \times h \] Here, \( m = 0.2 \mathrm{~kg}, \ g = 10 \mathrm{~m/s}^2, \ h = 2.2 \mathrm{~m} \).
03

Calculate Potential Energy Change

Calculate the change in potential energy when the ball rises to the maximum height. \[ \Delta U = 0.2 \mathrm{~kg} \times 10 \mathrm{~m/s}^2 \times 2.2 \mathrm{~m} = 4.4 \mathrm{~J} \] This potential energy change corresponds to the amount of work performed by the applied force.
04

Calculate Applied Force

Equation for work done by the force is \[ F \times 0.2 \mathrm{~m} = 4.4 \mathrm{~J} \] Solving for \( F \), we have \[ F = \frac{4.4 \mathrm{~J}}{0.2 \mathrm{~m}} = 22 \mathrm{~N} \]
05

Verify Against Possible Answers

Check our computed force against the choices offered in the problem: a) 16 N, b) 20 N, c) 22 N, d) 44 N. The force we calculated, 22 N, matches option c.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy
When an object is moved within a gravitational field, it gains or loses potential energy. This energy is related to its position in the field and depends on three factors:
  • The mass of the object (\( m \)).
  • The gravitational acceleration (\( g \)).
  • The height (\( h \)) to which the object is moved.
Potential energy can be calculated using the formula: \[ \Delta U = m \cdot g \cdot h \]In this exercise, as the ball ascends to 2 meters, it gains potential energy. The problem specifies the ball's mass as 0.2 kg. Substitute in the values for mass, gravitational acceleration, and height to calculate the potential energy change: \(\Delta U = 0.2 \cdot 10 \cdot 2.2 = 4.4 \text{ J} \). The unit "J" stands for joules, which is a measure of energy. This calculation shows the energy required for the ball to reach its height, connecting it to the overall energy analysis in the problem.
Work Done
The concept of work done relates to how much energy is transferred by a force moving an object through a distance. In physics, work is done when a force causes an object to move. The formula is:
\[ W = F \cdot d \]
  • \( F \) is the force applied.
  • \( d \) is the distance over which the force acts.
For the ball in this scenario, work done by the applied hand force is what causes it to initially rise. This work transfers energy to the ball, which is then converted into potential energy as it moves higher. In this problem, the work done must equal the change in potential energy (4.4 J) since that energy is directly transferred to change its height. That relationship helps us bridge understanding between force, distance, and energy.
Force Calculation
To determine the force applied by the hand, we use the relationship between work done and potential energy change. Given that the work done equals the potential energy gained, you set the work done equation equal to the potential energy:
\[ F \cdot 0.2 = 4.4 \]To find the force (\( F \)), rearrange the equation:
\[ F = \frac{4.4}{0.2} \]Calculating this gives:\(F = 22 \text{ N} \). This answer means the force applied is 22 newtons. The calculation confirms that option (c) in the problem's multiple-choice selections is the correct one. It highlights how understanding fundamental principles like work and energy can solve real-world questions about movement and force. This approach is critical in mechanics and numerous practical applications.

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Most popular questions from this chapter

A body with mass \(5 \mathrm{~kg}\) is acted upon by a force \(\mathbf{F}=(-3 \hat{\mathbf{i}}+4 \hat{j}) \mathrm{N}\). If its initial velocity at \(t=0\) is \(\mathbf{v}=(6 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}) \mathrm{ms}^{-1}\), the time at which it will just have a velocity along the \(y\)-axis is (a) never (b) \(10 \mathrm{~s}\) (c) \(2 \mathrm{~s}\) (d) \(15 \mathrm{~s}\)

A solid disc of mass \(M\) is just held in air horizontal by throwing 40 stones per sec vertically upwards to strike the disc each with a velocity \(6 \mathrm{~ms}^{-1}\). If the mass of each stone is \(0.05 \mathrm{~kg}\). What is the mass of the disc? \(\left(g=10 \mathrm{~ms}^{-2}\right)\) (a) \(1.2 \mathrm{~kg}\) (b) \(0.5 \mathrm{~kg}\) (c) \(20 \mathrm{~kg}\) (d) \(3 \mathrm{~kg}\)

A cricket ball of mass 150 g collides straight with a bat with a velocity of \(10 \mathrm{~ms}^{-1}\). Batsman hits it straight back with a velocity of \(20 \mathrm{~ms}^{-1}\), If ball remains in contact with bat for \(0.18\), then average force exerted by the bat on the ball is (a) \(15 \mathrm{~N}\) (b) \(45 \mathrm{~N}\) (c) \(150 \mathrm{~N}\) (d) \(4.5 \mathrm{~N}\)

Two masses \(M\) and \(M / 2\) arejoined together by means of light inextensible string pasaed over a frictionless pulley as shown in figure. When the bigger mass is released, the small one will ascend with an acceleration of [Kerala CET 2005] (a) \(\frac{g}{3}\) (b) \(\frac{3 g}{2}\) (c) \(g\) (d) \(\frac{g}{2}\)

A body of mass \(2 \mathrm{~kg}\) travels according to law \(x(t)=p t+q t^{2}+r t^{3}\), where \(p=3 \mathrm{~ms}^{-1}, q=4 \mathrm{~ms}^{-1}\) and \(r=5 \mathrm{~ms}^{-3}\). The force acting on the body at \(t=2 \mathrm{~s}\) is [NCFRT Exemplat] (a) \(136 \mathrm{~N}\) (b) \(134 \mathrm{~N}\) (c) \(158 \mathrm{~N}\) (d) \(68 \mathrm{~N}\)
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