How Much Work Does a Man Perform by Jumping Off a Stationary Boat?

  • Thread starter Thread starter Karol
  • Start date Start date
  • Tags Tags
    Boat Rest
Click For Summary
SUMMARY

The discussion focuses on calculating the work performed by a man jumping off a stationary boat, utilizing the principles of conservation of momentum and kinetic energy. The man of mass m jumps to the left, imparting a velocity v to the boat of mass M. The derived equations show that the man's velocity after the jump is given by vm = (M/m)v, leading to the total kinetic energy E = (1/2)((M + (M^2/m))v^2). The participants express uncertainty about the correctness of the calculations, particularly regarding the kinetic energy formula.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Familiarity with kinetic energy equations
  • Basic knowledge of mass and velocity relationships
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Review the principles of conservation of momentum in closed systems
  • Study kinetic energy calculations in physics
  • Explore the implications of mass and velocity changes in dynamic systems
  • Practice solving problems involving momentum and energy conservation
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of momentum and energy conservation principles in action.

Karol
Messages
1,380
Reaction score
22

Homework Statement


A man of mass m is on a still boat of mass M. the man jumps to the left so he has only horizontal velocity.
Immediately afterwards the boat has velocity v.
What's the work the man made.

Homework Equations


Conservation of momentum: ##m_1v_1=m_2v_2##
Kinetic energy: ##E=\frac{1}{2}mv^2##

The Attempt at a Solution


The man's velocity, vm, in an inertial frame, after the jump:
$$Mv=mv_m\rightarrow v_m=\frac{M}{m}v$$
$$E=\frac{1}{2}\left[ Mv^2+\frac{M}{m}v^2 \right]=\frac{1}{2}\frac{M(m+1)}{m}v^2$$
I don't think it's the right answer
 
Physics news on Phys.org
What is the square of ##\frac{M}{m}v##?
 
  • Like
Likes   Reactions: Karol
$$E=\frac{1}{2}\left[ Mv^2+m\left( \frac{M}{m}v\right)^2 \right]=\frac{1}{2}\left( M+\frac{M^2}{m} \right)v^2$$

Thanks
 
  • Like
Likes   Reactions: DEvens

Similar threads

Replies
31
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
990
Replies
335
Views
17K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K