Average force in stopping a block

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Homework Help Overview

The discussion revolves around a physics problem involving a 300kg block suspended by a cable, which falls and comes to rest after impacting the ground. Participants explore concepts related to weight, gravitational potential energy, and average force during the stopping process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to calculate the weight of the block and its potential energy during the fall. Questions arise regarding the relationship between work done, force, and displacement. Some participants explore how to calculate average force based on work and distance.

Discussion Status

The discussion includes various calculations and interpretations of the problem. Some participants provide guidance on including units in calculations, while others question the correctness of previous calculations and assumptions about force and energy. Multiple interpretations of the problem are being explored without explicit consensus.

Contextual Notes

Participants are navigating through the implications of the problem's setup, including the distance the block travels while coming to rest and the definitions of work and force. There is an emphasis on understanding units and their relevance in calculations.

ride4life
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Homework Statement


A 300kg block is suspended by a cable.
a) what is the weight of the motor block?
b) the cable snaps and the block falls 3 metres to the ground below.
i) calculate the time taken for the block to reach the ground.
ii) calculate the impact velocity of the block with the ground
c) if the block comes to rest having caused a 1cm dent in the block (ie traveled 1cm in coming to rest), calculate the average force exerted by the ground in stopping the block.

Homework Equations


a) weight = mass x gravity
b) i) s = ut + 0.5at^2
ii) v = u + at
c) ?


The Attempt at a Solution


a) weight = 300x9.81 = 2943N down
b) i) 3 = 0.5 x 9.81 x t^2 t = 0.78s
ii) v = 9.81 x 0.78 = 7.65m/s down
c) no idea
 
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Work done against the block = change in gravitational potential energy of the block.
(Conservation of energy)
 
work done = potential energy
w = m.g.h
w = 300x9.81x3
w = 8829 joules

is that right?
 
ride4life said:
work done = potential energy
w = m.g.h
w = 300x9.81x3
w = 8829 joules

is that right?

Right, so if you have the work done on the block. How would you find the force if you are given the distance?
 
work = force x displacement

foce = 8829/3

force = 2943N
 
ride4life said:
work = force x displacement

foce = 8829/3

force = 2943N

The block travels 1cm before coming to rest.
 
2943 x 0.01?
 
ride4life said:
2943 x 0.01?

the unit of 2943 x 0.01 is not Newton (the question is about force)

rock.freak667 talked about what the force is if the block traveled 1 cm with some amount of work done.
 
so it's 2943x0.01 J?
 
  • #10
ride4life said:
calculate the average force exerted by the ground in stopping the block.

What is the unit of force?
 
  • #11
songoku said:
What is the unit of force?

Newtons
 
  • #12
ride4life said:
2943 x 0.01?

will the unit of this calculation be Newton?
 
  • #13
songoku said:
will the unit of this calculation be Newton?

Newton seconds?
 
  • #14
lol

yes it's Newton seconds and that's unit of momentum, not force.
Since the question is about force, no doubt the answer is wrong.

you have the distance and work, find F
 
  • #15
s = 0.01m
W = 300x9.81x3
F = W/s
F = 882900N
 
  • #16
Looks good.

I recommend getting in the habit of including units during your calculations, and not just sticking them on at the end. Including units should help you understand better what's going on, and will help you catch errors like thinking that "2943 x 0.01" represents a force.
 
  • #17
Redbelly98 said:
Looks good.

I recommend getting in the habit of including units during your calculations, and not just sticking them on at the end. Including units should help you understand better what's going on, and will help you catch errors like thinking that "2943 x 0.01" represents a force.

Ok, thanks for the advice! :biggrin:
 

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