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Chapter 8: Problem 32

A particle of mass \(m\) is projected at an angle a to the horizontal with an initial velocity \(u\). The work done by gravity during the time it reaches its highest point is (1) \(u^{2} \sin ^{2} \alpha\) (2) \(\frac{m u^{2} \cos ^{2} \alpha}{2}\) (3) \(\frac{m u^{2} \sin ^{2} \alpha}{2}\) (4) \(-\frac{m u^{2} \sin ^{2} \alpha}{2}\)

Short Answer

Expert verified
The correct option is (4) \(-\frac{mu^2\sin^2\alpha}{2}\).

Step by step solution

01

Understand the Problem

A particle of mass \(m\) is projected into the air with an initial velocity \(u\) at an angle \(\alpha\) to the horizontal, reaching a highest point in its trajectory. Our task is to find the work done by gravity during this ascent.
02

Identify Key Concepts

Work done by gravity is the change in gravitational potential energy during the ascent to the highest point. At the highest point, the vertical component of velocity becomes zero.
03

Calculate Initial Vertical Velocity

The vertical component of the velocity \(u_y\) is \(u\sin\alpha\). This determines how high the particle can ascend before gravity brings it to a temporary stop (highest point).
04

Use Energy Concepts to Find Maximum Height

The kinetic energy associated with the vertical velocity at launch is \(\frac{1}{2}mu^2\sin^2\alpha\). At the highest point, all of this energy is converted into potential energy, which equals the work done by gravity.
05

Calculate Work Done by Gravity

Work done by gravity is equal to the negative change in potential energy. Since it was initially kinetic (vertically), the work done by gravity is \[\text{work done by gravity} = -\frac{mu^2\sin^2\alpha}{2}.\]
06

Choose the Correct Option

Comparing our result with the given options, the work done by gravity is, Option (4) \(-\frac{mu^2\sin^2\alpha}{2}\).

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Key Concepts

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

Work Done by Gravity
Whenever an object is moved in a gravitational field, gravity can do work on it. In the context of the projectile, when the particle is launched upwards, gravity performs negative work because it acts in the opposite direction to the direction of motion.
  • **Gravity's Work:** It is calculated as the product of force (gravity) and the displacement in the direction of force.
  • **Potential Energy Context:** The work done by gravity is equivalent to the change in gravitational potential energy as the object reaches its highest point.
For the projectile at hand, as it ascends, gravity pulls it back, converting kinetic energy into potential energy. The work done by gravity can be quantified as the difference in gravitational potential energy from the launch point to the highest point, which is \[- rac{mu^2 ext{sin}^2 ext{伪}}{2}.\]
Kinematic Equations
Kinematic equations are essential tools for solving motion-related problems, especially for projectiles. They relate variables such as velocity, acceleration, displacement, and time to describe motion precisely.
  • **Projectile Motion:** Analyzing motion involves breaking it into horizontal and vertical components.
  • **Main Equations:** The vertical motion of a projectile is commonly described using: - \[ v_y = u_y + at \] - \[ s = u_yt + \frac{1}{2}at^2 \] - \[ v_y^2 = u_y^2 + 2as \]
In our scenario, gravity directly affects the vertical motion. When the projectile reaches its highest point, its vertical velocity becomes zero, marking the temporary cessation of upward motion before gravity takes over completely.
Vertical Component of Velocity
The vertical component of velocity is a crucial part of projectile motion, affecting how high and how long an object will travel.
  • **Initial Vertical Velocity:** Often calculated as \( u_y = u ext{sin} ext{伪} \), it determines the upward force and height achieved.
  • **Vertical Motion Dynamics:** Influenced by gravity, constantly reducing upward speed until the apex is met. - At this point, the vertical velocity is zero, and the object has reached its highest point. - Understanding this change is key to solving for the work done by gravity.
Recognizing how the vertical component changes over time allows for solving dynamics of projectile motion effectively, where gravity is the sole influencing force after launch.

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Most popular questions from this chapter

A single conservative force \(F(x)\) acts on a \(1.0-\mathrm{kg}\) particle that moves along the \(x\)-axis. The potential energy \(U(x)\) is given by \(U(x)=20+(x-2)^{2}\) where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\), the particle has a kinetic energy of \(20 \mathrm{~J}\). Determine the equation of \(F(x)\) as a function of \(x\). (1) \(F=2+x\) (2) \(F=2+3 x\) (3) \(F=2(2-x)\) (4) \(F=3+2 x\)

A rope ladder of length \(L\) is attached to a balloon of mass \(M\). As the man of mass \(m\) climbs the ladder into the balloon basket, the balloon comes down by a vertical distance s. Then the increase in potential energy of man divided by the increase in potential energy of balloon is (1) \(\frac{L-s}{s}\) (2) \(\frac{L}{s}\) (3) \(\frac{s}{L-s}\) (4) \(L-s\)

A toy gun uses a spring of force constant \(K\). Before being triggered in the upward direction, the spring is compressed by a distance \(x .\) If the mass of the shot is \(m\), on being triggered, it will go up to a maximum height of (1) \(\frac{K x^{2}}{m g}\) (2) \(\frac{x^{2}}{K m g}\) (3) \(\frac{K x^{2}}{2 m g}\) (4) \(\frac{K^{2} x^{2}}{m g}\)

A ball of mass \(m\) is attached to the lower end of light vertical spring of force constant \(k\). The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstretched) length and comes to rest again after descending through a distance \(x\). (1) \(x=m g / k\) (2) \(x=2 m g / k\) (3) The ball will have no acceleration at the position where it has descended through \(x / 2\). (4) The ball will have an upward acceleration equal to \(g\) at its lowermost position

A force \(\vec{F}=(3 x y-5 z) \hat{j}+4 z \hat{k}\) is applied on a particle. The work done by the force when the particle moves from point \((0,0,0)\) to point \((2,4,0)\) as shown in figure is (1) \(\frac{280}{5}\) units (2) \(\frac{140}{5}\) units (3) \(\frac{232}{5}\) units (4) \(\frac{192}{5}\) units
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