Calculating Work Done on Child & Sled Pulled by Force

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SUMMARY

The discussion focuses on calculating the total work done on a child and a sled being pulled by a force of 100.00 N at a 45-degree angle, with a friction force of 30.0 N. The work done by the pulling force is calculated using the formula W = F * cos(theta) * d, resulting in 707 J. To find the total work done, the work against friction must also be considered, and the final speed can be determined using the net work done with the equation W = (1/2)mv².

PREREQUISITES
  • Understanding of basic physics concepts such as work and energy.
  • Familiarity with trigonometric functions, specifically cosine.
  • Knowledge of Newton's laws of motion.
  • Ability to manipulate algebraic equations for solving for variables.
NEXT STEPS
  • Learn about calculating work done against friction in physics.
  • Study the relationship between work and kinetic energy using the work-energy theorem.
  • Explore the concept of net force and its impact on motion.
  • Review trigonometric applications in physics, particularly in force analysis.
USEFUL FOR

This discussion is beneficial for students studying physics, particularly those new to concepts of work, energy, and forces. It is also useful for anyone needing to apply these principles in practical scenarios, such as in engineering or physical education.

fa08ti
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A child on a sled (having a combined mass of 47.0 kg) is pulled by a force directed along a rope that makes a 45 degree angle with the horizontal axis. the force exerted on the rope is 100.00 N. the force of friction acting on the sled is 30.0 N. if the child is pulled a distance of 10.0 m along a level field, determine the total work done on the child and the sled.

Attempt:
Given:
d: 10.0 m, F:100.0 N (45 degrees), Ff: 30.0 N (180 Degrees)

W=F(costheta) . deltad
W= 100.00 N X cos45 X 10.0m
W=707J
that would be just for the rope

would i have to do the same calculation using the Ff now? then add both numbers to get total work done?

It also asks for the final speed at the end of 10.0 m

HELP PLEASE!
 
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fa08ti said:
would i have to do the same calculation using the Ff now? then add both numbers to get total work done?
Yes.

It also asks for the final speed at the end of 10.0 m
Make use of the net work done that you've calculated.
 
ok so for the speed, i'd use
W= (1/2)mv^2 and just rearrange it
 
Last edited:
fa08ti said:
ok so for the speed, i'd use
W= (1/2)mv^2 and just rearrange it
Good!
 
thanks sooooo much. I've never taken physics before and I'm finding that while i understand most concepts, I tend to need a lot of reassurance
 

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