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DiaperBoy35
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Physics question on Energy and Spring, any help appreciated :D
a. A 1.0 kg puck sliding at 15 m/s along some horizontal frictionless ice strikes and compresses a horizontal spring attached to one end of the ice rink. If the spring has a constant of 35 N/m, what is the maximum compression of the spring?
b. If in the above problem, the puck experiences a constant frictional force of 4.0 N opposing its motion beginning when it first strikes the spring, what would the maximum compression of the spring now be? (10 marks)
I am unsure about part b, as i am not sure what to use for the value of Δd as you will see underneath at my attempt
-0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1)
- E(thermal) = Frictional force x Δd
E(total 1) = E (total 2)
0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1) [ 0.5kv^2 (1) and 0.5mv^2 (1) equal 0]
0.5mv^2 (1) = 0.5kv^2 (1)
x = √((15^2)/35)
x = 2.535 m
[ (1) <-- denotes to initial (2)<-- denotes to final]
Part b
E(total 1) = E (total 2)
0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1) + E(thermal) [ 0.5kv^2 (1) and 0.5mv^2 (1) equal 0]
0.5mv^2 (1) = + 0.5kv^2 (1) + Ff Δ d
I am unsure of what to do from here
[ (1) <-- denotes to initial (2)<-- denotes to final]
Homework Statement
a. A 1.0 kg puck sliding at 15 m/s along some horizontal frictionless ice strikes and compresses a horizontal spring attached to one end of the ice rink. If the spring has a constant of 35 N/m, what is the maximum compression of the spring?
b. If in the above problem, the puck experiences a constant frictional force of 4.0 N opposing its motion beginning when it first strikes the spring, what would the maximum compression of the spring now be? (10 marks)
I am unsure about part b, as i am not sure what to use for the value of Δd as you will see underneath at my attempt
Homework Equations
-0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1)
- E(thermal) = Frictional force x Δd
The Attempt at a Solution
E(total 1) = E (total 2)
0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1) [ 0.5kv^2 (1) and 0.5mv^2 (1) equal 0]
0.5mv^2 (1) = 0.5kv^2 (1)
x = √((15^2)/35)
x = 2.535 m
[ (1) <-- denotes to initial (2)<-- denotes to final]
Part b
E(total 1) = E (total 2)
0.5mv^2 (1) + 0.5kv^2 (1)= 0.5mv^2 (1) + 0.5kv^2 (1) + E(thermal) [ 0.5kv^2 (1) and 0.5mv^2 (1) equal 0]
0.5mv^2 (1) = + 0.5kv^2 (1) + Ff Δ d
I am unsure of what to do from here
[ (1) <-- denotes to initial (2)<-- denotes to final]
