WORK by Constant Force Argh help

In summary, a freight company uses a compressed spring to shoot packages up a frictionless ramp into a truck. The spring constant is 337 N/m and the spring is compressed 34.0 cm. For part A, the speed of the package when it reaches the truck can be calculated using conservation of energy. However, for part B, a sticky spot with a coefficient of kinetic friction 0.30 creates a potential obstacle for the next package to make it into the truck. To solve for part A, we need to consider the energy when the spring is fully compressed and the package is at its lowest point. For part B, we need to use conservation of energy to find out how much kinetic energy the package has when it reaches the
  • #1
Kalie
46
0
WORK by Constant Force...Argh...help

A freight company uses a compressed spring to shoot 1.80 kg packages up a 1.0-m-high frictionless ramp into a truck, as the figure shows. The spring constant is 337 N/m and the spring is compressed 34.0 cm.

A) What is the speed of the package when it reaches the truck?

B) A careless worker spills his soda on the ramp. This creates a 50-cm-long sticky spot with a coefficient of kinetic friction 0.30. Will the next package make it into the truck? Yes or No?

Okay...For Part A this is what I did but it was wrong

First I said F_s=-k x= 337(.34) = 114.38
with this I said that 114.38=1/2mv^2 v=11.28
The F_w=1/2mv^2=ma=1.8*-9.8=-17.64
W=F_w*D=F_w=17.64
W= the change in KE
KE_f=W+KE_i
KE_f=96.74
1/2mv^2=96.74
so v=10.36
That is wrong and I don't know how to get the right answer could someone help?

Part B will probably make more sense after I solve the first part but could someone help me with the setup?
 
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  • #2
Kalie said:
Okay...For Part A this is what I did but it was wrong

First I said F_s=-k x= 337(.34) = 114.38
Here you calculate the force of the spring when it is at max compression.
with this I said that 114.38=1/2mv^2 v=11.28
Then you set this force equal to the KE! :yuck: (Force and energy are two different things!)
The F_w=1/2mv^2=ma=1.8*-9.8=-17.64
W=F_w*D=F_w=17.64
W= the change in KE
KE_f=W+KE_i
KE_f=96.74
1/2mv^2=96.74
so v=10.36
That is wrong and I don't know how to get the right answer could
Try this: Consider conservation of energy. What's the energy when the spring is fully compressed and the package is at it's lowest point? (You didn't supply the figure, so it's not clear how the spring and ramp are connected. Is the spring on the ramp? Post the figure if you can.)
 
  • #4
OK. Now use conservation of energy to find out how much KE the package has when it gets to the top of the ramp. Hint: Consider spring PE, gravitational PE, and KE.
 

What is "WORK by Constant Force Argh help"?

"WORK by Constant Force Argh help" is a scientific concept that describes the amount of energy required to move an object a certain distance when a constant force is applied to it. It is often used to measure the effort or exertion needed to perform a task.

How is "WORK by Constant Force Argh help" calculated?

The formula for calculating "WORK by Constant Force Argh help" is W = F * d, where W is the work done, F is the applied force, and d is the distance the object is moved. The unit of measurement for work is joules (J).

What is the difference between "WORK by Constant Force Argh help" and power?

While "WORK by Constant Force Argh help" measures the amount of energy used to move an object, power measures the rate at which work is done. In other words, power is the amount of work done per unit of time. The unit of measurement for power is watts (W).

How does the direction of the force affect "WORK by Constant Force Argh help"?

The direction of the applied force does not affect the amount of "WORK by Constant Force Argh help" done. The formula for calculating work only takes into account the magnitude of the force and the distance moved, not the direction of the force.

What are some real-life examples of "WORK by Constant Force Argh help"?

"WORK by Constant Force Argh help" can be seen in activities such as pushing a shopping cart, lifting weights, or pushing a lawn mower. It is also used in more complex tasks such as moving heavy objects with cranes or rockets launching into space.

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